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COMPLETE 


AEITHMETIC, 


ORAL  AKD    WEITTE]Sr. 


PART  SECOND, 


BY 

MALCOLM    MacVIOAR,    Ph.D.,    LL.D., 

FBINCIFAIi  STATB  KOBMAL  SCHOOL,  POTSDAM,  N.  Y. 


PUBLISHED   BY 

TAINTOR  BROTHERS,  MERRILL   &   CO., 

NEW    YORK. 


^ 


Copyright  by 
TAINTOR  BROTHERS,  MERRILL  &  CO., 

1878. 


Electrotyped  by  Smith  &  MoDougal,  83  Beekman  St.,  N.  Y. 


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♦-*s^|  PREFACE.  JJt^^^^ 


•B ■«r:y:77.CT'^ ff" 


^^^^ 

THE  aim  of  the  author  in  the  preparation  of  this  work  may 
be  stated  as  follows : 

1.  To  present  each  subject  in  arithmetic  in  such  a  manner 
as  to  lead  the  pupil  by  means  of  preparatory  steps  and  proposi- 
tions which  he  is  required  to  examine  for  himself,  to  gain  clear 
perceptions  of  the  elements  necessary  to  enable  him  to  grasp 
as  a  reality  the  more  complex  and  complete  processes. 

2.  To  present,  wherever  it  can  be  done,  each  process  object- 
ively, so  that  the  truth  under  discussion  is  exhibited  to  the  eye 
and  thus  sharply  defined  in  the  mind. 

3.  To  give  such  a  systematic  drill  on  oral  and  written  exer- 
cises and  review  and  test  questions  as  will  fix  permanently  in 
the  mind  the  principles  and  processes  of  numbers  with  their 
applications  in  practical  business. 

4.  To  arrange  the  pupil's  work  in  arithmetic  in  such  a  manr 
ner  that  he  will  not  fail  to  acquire  such  a  knowledge  of  princi- 
ples and  facts,  and  to  receive  such  mental  discipline,  as  will  fit 
him  properly  for  the  study  of  the  higher  mathematics. 

The  intelligent  and  experienced  teacher  can  readily  deter- 
mine by  an  examination  of  the  work  how  well  the  author  has 
succeeded  in  accomplishing  his  aim. 

Si5770ri7 


PREP  ACE. 

Special  attention  is  invited  to  the  method  of  presentation 
given  in  the  teacher's  edition.  This  is  arranged  at  the  begin- 
ning of  each  subject,  just  where  it  is  required,  and  contains 
definite  and  full  instructions  regarding  the  order  in  which  the 
subject  should  be  presented,  the  points  ^at  require  special 
attention  and  illustration,  the  kind  of  illustrations  that  should 
be  used,  a  method  for  drill  exercise,  additional  oral  exercises 
where  required  for  the  teacher's  use,  and  such  other  instructions 
as  are  necessary  to  form  a  complete  guide  to  the  teacher  in  the 
discussion  and  presentation  of  each  subject. 

The  plan  adopted  of  having  a  separate  teacher's  edition 
avoids  entirely  the  injurious  course  usually  pursued  of  cum- 
bering the  pupil's  book  with  hints  and  suggestions  which  are 
intended  strictly  for  the  teacher. 

Attention  is  also  invited  to  the  Properties  of  Numbers,  Great- 
est Common  Divisor,  Fractions,  Decimals,  Compound  Num- 
bers, Business  Arithmetic,  Ratio  and  Proportion,  Alligation, 
and  Square  and  Cube  Root,  with  the  belief  that  the  treatment 
will  be  found  new  and  an  improvement  upon  former  methods. 

The  author  acknowledges  with  pleasure  his  indebtedness  to 
Prof.  D.  H.  Mac  Vicar,  LL.D.,  Montreal,  for  valuable  aid 
rendered  in  the  preparation  of  the  work,  and  to  Charles  D. 
McLean,  A.  M.,  Principal  of  the  State  Normal  and  Training 
School,  at  Brockport,  N.  Y.,  for  valuable  suggestions  on 
several  subjects. 

M.  MacVICAR. 
Potsdam,  September ^  1877. 


l^r^i^r^ 


^^ 


/J^NS^ 


BEVIEW   OF   PART    FIBST. 

PAGE 

Notation  and  Numeration.  1 

Addition 4 

Subtraction 5 

Multiplication 7 

Division 11 

Properties  of  Numbers.  ...  16 

Exact  Division 16 

Prime  Numbers 19 

Factoring 20 

Cancellation 31 

Greatest  Common  Divisor..  22 

Least  Common  Multiple —  25 

Fractions 29 

Complex  Fractions 40 

Decimal  Fractions 46 

Denominate  Numbers 59 

Metric  System 75 

part  second, 

Business  Arithmetic 79 

Aliquot  Parts 82 

Business  Problems  85 

Applications 100 

Profit  and  Loss 101 

Commission 103 

Insurance 105 

Stocks 107 

Taxes Ill 

Duties  or  Customs 113 

Review  and  Test  Questions.  114 

Interest 115 


FAGS 

Method  by  Aliquot  Parts.  117 
Method  by  Six  Per  Cent..  119 

Method  by  Decimals 121 

Exact  Interest 122 

Compound  Interest 127 

Interest  Tables 120 

Annual  Interest 131 

Partial  Payments 1 32 

Discount 136 

Bank  Discount 138 

Exchange 141 

Domestic  Exchange 142 

Foreign  Exchange 146 

Equation  of  Payments 151 

Review  and  Test  Questions.  160 

Ratio 161 

Proportion 169 

Simple  Proportion 170 

Compound  Proportion 174 

Partnership 177 

Alligation  Medial 181 

Alligation  Alternate 182 

Involution 187 

Evolution 189 

Progressions 205 

Arithmetical  Progression.  206 
Geometrical  Progression..  208 

Annuities 210 

Mensuration 213 

Review  and  Test  Examples.  227 
Answers 243 


♦>♦»♦■ 


REVIEW  OF  PART  FIRST.* 


,<»wfi^^^S^^^'g)ij/ 


NOTATION   AND   NUMERATION, 


DEFINITIONS. 

11.  A  Unit  is  a  single  thing,  or  group  of  single  things, 
regarded  as  one  ;  as,  one  ox,  one  yard,  one  ten,  one  hundred. 

12.  Units  are  of  two  hinds  —  Mathematical  and 
Common.  A  mathematical  unit  is  a  single  thing  which  has  a 
fixed  value ;  as,  one  yard,  one  quart,  one  hour,  one  ten,  A 
common  unit  is  a  single  thing  which  has  no  fixed  value ;  as, 
one  house,  one  tree,  one  garden,  one  farm. 


*  Note. — The  first  78  pages  of  this  part  contains  so  much  of  the  matter 
in  Part  First  as  is  necessary  for  a  thorough  review  of  each  subject,  in- 
cluding all  the  tables  of  Compound  Numbers.  For  convenience  in 
making  references,  the  Articles  retained  are  numbered  the  same  as  in 
Part  First.     Hence  the  numbers  of  the  Articles  are  not  consecutive. 


2  NOTATION    AND     NUMERATION. 

13.  A  Number  is  a  unit,  or  collection  of  units ;  as,  one 
man,  three  houses,  four,  six  hundred. 

Observe,  the  number  is  "the  how  many"  and  is  represented  by  what- 
ever answers  the  question,  How  many  ?  Thus  in  the  expression  seven 
j&rdB,  seven  represents  the  number. 

14.  The  Unit  of  a  Number  is  one  of  the  things 
numbered. 

Thus,  the  unit  of  eight  bushels  is  one  bushel,  of  five  boys  is  one  boy, 
of  nine  is  one. 

15.  A  Concrete  Number  is  a  number  which  is  applied 
to  objects  that  are  named ;  as  four  chairs,  ten  hells. 

16.  An  Abstract  Number  is  a  number  which  is  not 
applied  to  any  named  objects ;  as  nine,  five,  thirteen. 

17.  Like  Numbers  are  such  as  have  the  same  unit. 
Thus,  four  windows  and  eleven  windows  are  like  numbers,  eight  and 

ten,  three  hundred  and  seven  hundred. 

18.  Unlike  Numbers  are  such  as  have  different  units. 
Thus,  twelve  yards  and  five  days  are  unlike  numbers,  also  six  cents 

and  nine  minutes, 

19.  Figures  are  characters  used  to  express  numbers. 

20.  The  Value  of  a  figure  is  the  number  whicii  it 
represents. 

21.  The  Simple  or  Absolute  Value  of  a  figure  is  the 
number  it  represents  when  standing  alone,  as  8. 

22.  The  Local  or  Representative  Value  of  a  figure 
is  the  number  it  represents  in  consequence  of  the  place  it 
occupies. 

Thus,  in  66  the  6  in  the  second  place  from  the  right  represents  a  num- 
ber ten  times  as  great  as  the  6  in  the  first  place. 

2.3.  Notation  is  the  method  of  writing  numbers  by 
means  of  figures  or  letters. 


NOTATION    AND     NUMERATION.  3 

24.  Numeratioii  is  the  method  of  reading  numbers 
which  are  expressed  by  figures  or  letters. 

35.  A  Scale  in  Arithmetic  is  a  succession  of  mathematical 
units  which  increase  or  decrease  in  value  according  to  a  fixed 
order. 

26.  A  Decimal  Scale  is  one  in  which  the  fixed  order 
of  increase  or  decrease  is  uniformly  ten. 

This  is  the  scale  used  in  expressing  numbers  by  figures. 

•  27.  Arithinetic  is  the  Science  of  Numbers  and  the  Art 
of  Computation. 

REVIEW    AND    TEST    QUESTIONS. 

31.  Study  carefully  and  answer  each  of  the  following 
questions : 

1.  Define  a  scale.     A  decimal  scale. 

2.  How  many  figures  are  required  to  express  numbers  in  the  decimal 
scale,  and  why  ? 

3.  Explain  the  use  of  the  cipher,  and  illustrate  by  examples, 

4.  State  reasons  why  a  scale  is  necessary  in  expressing  numbers. 

5.  Explain  the  use  of  each  of  the  three  elements— /^?/re."?,  place,  and 
comma — in  expressing  numbers. 

6.  What  is  meant  by  the  simple  or  absolute  value  of  figures  ?  What 
by  the  local  or  representative  value? 

7.  How  is  the  local  value  of  a  figure  affected  by  changing  it  from  the 
Jirst  to  the  tliird  place  in  a  number  ? 

8.  How  by  changing  a  figure  from  the  second  to  the  fourth  ?  From 
the  fourth  to  the  ninth  ? 

9.  Explain  how  the  names  of  numbers  from  twelve  to  twenty  are 
formed.     From  twenty  to  nine  hundred  ninety. 

10.  What  is  meant  by  a  period  of  figures  ? 

11.  Explain  how  the  name  for  each  order  in  any  period  is  formed. 

12.  State  the  name  of  the  right-hand  order  in  each  of  the  first  six 
periods,  commencing  with  units. 

13.  State  the  two  things  mentioned  in  (9)  which  must  be  observed 
when  writing  large  numbers. 

14.  Give  a  rule  for  reading  numbers ;  also  for  writing  numbers. 


4  ADDITION, 

ADDITIOlSr. 

50.  Addition  is  the  process  of  uniting  two  or  more 
numbers  into  one  number. 

51.  uMdends  are  the  numbers  added. 

52.  The  Sum  or  Amount  is  the  number  found  by 
addition. 

53.  The  Process  of  Addition  consists  in  forming 
units  of  the  same  order  into  groups  of  ten,  so  as  to  express 
their  amount  in  terms  of  a  higher  order. 

54.  The  Sign  of  Addition  is  +,  and  is  read  plus. 

When  placed  between  two  or  more  numbers,  thus,  8  +  3  +  6  +  3  +  9,  it 
means  that  they  are  to  be  added. 

55.  The  Sif/n  of  Equality  is  =,  and  is  read  equals, 
or  equal  to  ;  thus,  9  +  4  =  13  is  read,  nine  plus  four  equals 
thirteen, 

5G.  Principles. — /.  Only  numhers  of  the  same  denom- 
ination and  units  of  the  same  order  can  he  added. 

II.  The  sum  is  of  the  same  denomination  as  the 
addends. 

III.  The  whole  is  equal  to  the  sum  of  all  the  parts, 

REVIEW    AND    TEST    QUESTIONS. 

57.     1.  Define  Addition,  Addends,  and  Sum  or  Amount. 

2.  Name  each  step  in  the  process  of  Addition. 

3.  Why  place  the  numbers,  preparatory  to  adding,  units  under  units, 
tens  under  tens,  etc.? 

4.  Why  commence  adding  with  the  units'  column  ? 

5.  What  objections  to  adding  the  columns  in  an  irregular  order? 
Illustrate  by  an  example. 

6.  Construct,  and  explain  the  use  of  the  addition  table. 

7.  IIow  many  combinations  in  the  table,  and  how  found? 

8.  Explain  carrying  in  addition.  What  objection  to  the  use  of  the 
word? 


SUBTRACTION.  5 

9.  Define  counting  and  illustrate  by  an  example. 

10.  Write  five  examples  illustrating  the  general  problem  of  addition, 
"  Given  all  the  parts  to  find  the  whole." 

11.  State  the  diflerence  between  the  addition  of  objects  and  the  addi- 
tion  of  numbers. 

12.  Show  how  addition  is  performed  by  using  the  addition  table. 

13.  What  is  meant  by  the  denomination  of  a  number?  What  by 
units  of  the  same  order  ? 

14.  Show  by  analysis  that  in  adding  numbers  of  two  or  more  places, 
the  orders  are  treated  as  independent  of  each  other. 

SUBTEACTION. 

70,  Subtraction  is  the  process  of  finding  the  difference 
between  two  numbers. 

71«  The  Minuend  is  the  greater  of  two  numbers  whose 
difference  is  to  be  found. 

112 •  The  Subtrahend  is  the  smaller  of  two  numbers 
whose  difference  is  to  be  found. 

73.  The  Difference  or  Hetnainder  is  the  result 
obtained  by  subtraction. 

74.  The  Ih^ocess  of  Subtraction  consists  in  com- 
paring two  numbers,  and  resolving  the  greater  into  two  parts, 
one  of  which  is  equal  to  the  less  and  the  other  to  the  differ- 
ence of  the  numbers. 

75.  The  Sign  of  Subtraction  is  — ,  and  is  called 
minus. 

When  placed  between  two  numbers  it  indicates  that  their  difference 
is  to  be  found ;  thus,  14  —  6  is  read,  14  minus  6,  and  means  that  the  dif- 
ference between  14  and  6  is  to  be  found. 

76.  JParentheses  (  )  denote  that  the  numbers  inclosed 
between  them  are  to  be  considered  as  one  number. 

77.  A  Vinculum  affects  numbers  in  the  same 
manner  as  parentheses. 

Thus,  19 +  (13— 5),  or  19  +  13—5  signifies  that  the  difference  between 
13  and  5  is  to  b'e  added  to  19. 


6  SUBTRACTION. 

78.  Prtn'Ciples. — /.  Only  like  Jiurtihers  and  units  of 
the  same  order  ean  he  subtracted. 

II.  The  minuend  is  the  sum  of  the  subtrahend  and. 
difference,  or  the  minuend  is  the  whole  of  luhich  the 
subtrahend  and  difference  are  the  parts. 

III.  An  equal  increase  or  decrease  of  the  minuend 
and  subtrahend  does  not  change  the  difference. 

KEVIEW    AND    TEST    QUESTIONS. 

*79.  1.  Define  the  process  of  subtraction.  Illustrate  each  step  by 
an  example. 

2.  Explain  how  subtraction  should  be  performed  when  an  order  in  the 
subtrahend  is  greater  than  the  corresponding  order  in  the  minuend. 
Illustrate  by  an  example. 

3.  Indicate  the  difference  between  the  subtraction  of  numbers  and  the 
subtraction  of  objects. 

4.  When  is  the  result  in  subtraction  a  remainder,  and  when  a  differ- 
ence? 

5.  Show  that  so  far  as  the  process  with  numbers  is  concerned,  the 
result  is  always  a  difference. 

6.  Prepare  four  original  examples  under  each  of  the  following  prob- 
lems and  explain  the  method  of  solution  : 

Prob.  I.— Given  the  whole  and  one  of  the  parts  to  find  the  other  part. 
Prob.  II. — Given  the  sum  of  four  numbers  and  three  of  them  to  find 
the  fourth. 

7.  Construct  a  Subtraction  Table. 

8.  Define  counting  by  subtraction. 

9.  Show  that  counting  by  addition,  when  we  add  a  number  larger 
than  one,  necessarily  involves  counting  by  subtraction. 

10.  What  is  the  difference  between  the  meaning  of  denomination  and 
orders  of  units  ? 

11.  State  Principle  III  and  illustrate  its  meaning  by  an  example. 

12.  Show  that  the  difference  between  63  and  9  is  the  same  as  the 
difference  between  (63  +  10)  and  (9  +  10). 

13.  Show  that  28  can  be  subtracted  from  92,  without  analyzing  the 
minuend  as  in  (64),  by  adding  10  to  each  number. 

14.  What  must  be  added  to  each  number,  to  subtract  275  from  829 
without  analyzing  the  minuend  as  in  ((>4:)? 

15.  What  is  meant  by  borrowing  and  canning  in  subtraction  ? 


MULTIPLICATION.  7 

MULTIPLIOATIOIT. 

IZIjUSTMATION     of     l*JtOCESS. 

92.  Step  II. — To  multiply  by  using  the  parts  of  the 
multiplier. 

1.  The  multiplier  may  be  made  into  any  desired  parts,  and  tlie  mul- 
tiplicand taken  separately  the  number  of  times  expressed  by  each  part. 
The  sum  of  the  products  thus  found  is  the  required  product. 

Thus,  to  find  9  times  12  we  may  take  4  times  13  which  are  48,  then  5 
times  12  which  are  60.  4  times  12  plus  5  times  13  are  9  times  13 ; 
hence,  48  plus  60,  or  108,  are  9  times  13. 

2.  When  we  multiply  by  one  of  the  equal  parts  of  the  multiplier,  we 
find  one  of  the  equal  parts  of  the  required  product.  Hence,  by  multi- 
plying the  part  thus  found  by  the  number  of  such  parts,  we  find  the 
required  product. 

For  example,  to  find  12  times  64  we  may  proceed  thus : 


(1.)    ANALYSIS.  (2.) 

64x4  =  256^  64 

64x     4.  =  26e>=d  times  256       4 

64x4  =  256  3  256 

64x12  =  768  3 

768 


(1.)  Observe,  that  12  =  4  -f-  4  -f-  4  ;  hence,  4  is  one  of  the  3  equal 
parts  of  12. 

(2.)  That  64  is  taken  12  times  by  taking  it  4  times -I- 4  times -^  4  times, 
as  shown  in  the  analysis. 

(3.)  That  4  times  64,  or  256,  is  one  of  the  3  equal  parts  of  12  times  64. 
Hence,  multiplying  256  by  3  gives  12  times  64,  or  768. 

3.  In  multiplying  by  20,  30,  and  so  on  up  to  90,  we  invariably  multi- 
ply by  10,  one  of  the  equal  parts  of  these  numbers,  and  then  by  the 
number  of  such  parts. 

For  example,  to  multiply  43  by  30,  we  take  10  times  43,  or  430,  and 
multiply  this  product  by  3 ;  430  x  3  =  1290,  which  is  30  times  43.  We 
multiply  in  the  same  manner  by  300,  300,  etc.,  2000,  3000,  etc. 


f  MULTIPLICATION. 

93.  Prob.  II. — To  multiply  by  a  number  containing 
only  one  order  of  units. 

1.  Multiply  347  by  500. 

(1.)    ANALYSIS.  (2.) 

Firststep,        347x100=     34700  347 

Second  step,      34700   X  5  =  173500  500 

173500 

Explanation. — 500  is  equal  to  5  times  100 ;  hence,  by  taking  347, 
as  in  first  step,  100  times,  5  times  this  result,  or  5  times  34700,  as  shown 
in  second  step,  will  make  500  times  347.  Hence  173500  is  500 
times  347. 

In  practice  we  multiply  first  by  the  significant  figure,  and  annex  to 
the  product  as  many  ciphers  as  there  are  ciphers  in  the  multiplier,  as 
shown  in  (2). 

96.  Prob.  III. — To  multiply  by  a  number  containing 
two  or  more  orders  of  units. 

1.  Multiply  539  by  374. 

(1.)  analysis.  (2.) 


-~\ 


5  3  9     Multiplicand. 
3  7  4     Multiplier. 


539X  4=  2156  Ist  partial  product. 

539X374=<5539X       70=      37  7  30  2d  partial  product. 

539x300  =  161700  3d  partial  product. 

2  015  8  6  Whole  product. 


Explanation. — 1.  The  multiplier,  874,  is  analyzed  into  the  parts  4. 
70,  and  300,  according  to  (92). 

2.  The  multiplicand,  539,  is  taken  first  4  times  =  2156  (86) ;  then 
70  times  =  37730  (93) ;  then  300  times  =  161700  (93). 

3.  4  times  +  70  times  +  300  times  are  equal  to  374  times  ;  hence  the 
sum  of  the  partial  products,  2156,  37780,  and  161700,  is  equal  to  374 
times  539  =  201586. 

4.  Observe,  that  in  practice  we  arrange  the  partial  products  as  shown 
in  (2),  omitting  the  ciphers  at  the  right,  and  placing  the  first  significant 
figure  of  each  product  under  the  order  to  which  it  belongs. 


MULTIPLICATION.  SF 


DEFINITIONS. 

100.  Multiplication  is  the  process  of  taking  one 
number  as  many  times  as  there  are  units  in  another. 

101.  The  Multiplicand  is  the  number  taken,  or  mul- 
tiplied. 

102.  The  Multiplier  is  the  number  which  denotes  how 
many  times  the  multiplicand  is  taken. 

103.  The  I^roduct  is  the  result  obtained  by  multipli- 
cation. 

104.  A  Partial  Product  is  the  result  obtained  by 
multiplying  by  one  order  of  units  in  the  multiplier,  or  by  any 
part  of  the  multiplier. 

105.  The  Total  or  Whole  Product  is  the  sum  of  all 

the  partial  products. 

106.  The  Process  of  Multiplication  consists, /rs^, 
in  finding  partial  products  by  using  the  memorized  results  of 
the  Multiplication  Table;  second,  in  uniting  these  partial 
products  by  addition  into  a  total  product. 

107.  A  Factor  is  one  of  the  equal  parts  of  a  number. 

Thus,  13  is  composed  of  six  2's,  four  3's,  three  4's,  or  two  6's  ;  hence, 
2,  3,  4,  and  6  are  factors  of  12. 

The  multiplicand  and  multiplier  are  factors  of  the  product.  Thus, 
37  X  25  =  925.  The  product  925  is  composed  of  twenty-five  37's,  or 
thirtyse'oen  25's.  Hence,  both  37  and  25  are  equal  parts  or  factors 
of  925. 

108.  The  Sign  of  Multiplication  is  x,  and  is  read 
times,  or  miiUipUed  hy. 

When  placed  between  two  numbers,  it  denotes  that  either  is  to  be 
multiplied  by  the  other.  Thus,  8x6  shows  that  8  is  to  be  taken  6  times, 
or  that  6  is  to  be  taken  8  times  ;  hence  it  may  be  read  either  8  times  6  or 
6  times  8. 


10  MULTIPLICATION. 

109.  Principles. — /.  Tlxe  multiplicand  may  he  either 
an  abstract  or  concrete  nunvber. 

II.  The  multiplier  is  always  an  abstract  number. 

III.  Tize  product  is  of  the  same  denomination  as  the 
multiplicand. 

KEVIEW  AND  TEST  QUESTIONS. 

110.  1.  Define  Multiplication,  Multiplicand,  Multiplier,  and 
Product. 

2.  What  is  meant  by  Partial  Product  ?     Illustrate  by  an  example. 

3.  Define  Factor,  and  illustrate  by  examples. 

4.  What  are  the  factors  of  6  ?  14  ?  15  ?  9  ?  20  ?  24  ?  25  ?  27  ? 
32?    10?    30?    50?    and    70? 

5.  Show  that  the  multiplicand  and  multiplier  are  factors  of  the 
product. 

6.  What  must  the  denomination  of  the  product  always  be,  and  why  ? 

7.  Explain  the  process  in  each  of  the  following  cases  and  illustrate 
by  examples : 

I.  To  multiply  by  numbers  less  than  10. 

II.  To  multiply  by  10,  100,  1000,  and  so  on. 

III.  To  multiply  by  one  order  of  units. 

XV.  To  multiply  by  two  or  more  orders  of  units. 

V.  To  multiply  by  the  factors  of  a  number  (92 — ^2).    . 

8.  Give  a  rule  for  the  third,  fourth,  and  fifth  cases. 

9.  Give  a  rule  for  the  shortest  method  of-  working  examples  where 
both  the  multiplicand  and  multiplier  have  one  or  more  ciphers  on 
the  right. 

10.  Show  how  multiplication  may  be  performed  by  addition. 

11.  Explain  the  construction  of  the  Multiplication  Table,  and  illus- 
trate its  use  in  multiplying. 

12.  Why  may  the  ciphers  be  omitted  at  the  right  of  partial 
products  ? 

13.  Why  commence  multiplying  the  units'  order  in  the  multi- 
plicand first,  then  the  tens',  and  so  on?  Illustrate  your  answer  by 
an  example. 

14.  Multiply  8795  by  629,  multiplying  first  by  the  tens,  then  by  the 
hundreds,  and  last  by  the  units. 


DIVISION.  11 

15.  Multiply  3573  by  483,  commencing  with  the  thousands  of  the 
multiplicand  and  hundreds  of  the  multiplier. 

16.  Show  that  hundreds  multiplied  by  hundreds  will  give  ten  thou- 
sands \u.  the  product. 

17.  Multiplying  thousands  by  thousands,  what  order  will  the  pro- 
duct be? 

18.  Name  at  sight  the  lowest  order  which  each  of  the  following 
examples  will  give  in  the  product  : 

(1.)  8000  X  8000 ;  2000000  x  3000 ;  5000000000  x  7000. 
(2.)  40000  X  20000 ;  7000000  x  4000000. 

19.  What  orders  in  3928  can  be  multiplied  by  each  order  in  473,  and 
not  have  any  order  in  the  product  less  than  thousands  ? 


DIVISION. 

ILLUSTRATION     OF     PROCESS. 

119.  Prob.  I. — To  divide  any  number  by  any  divisor 
not  greater  tlian  12. 

1.  Divide  986  by  4. 

Explanation. — Follow  the  analysis  and  notice  each  step  in  the 
process;  thus, 

1.  We  commence  by  dividing  the  higher  order  of  units.  We  know 
that  9,  the  figure  expressing  hundreds,  contains  twice  the  divisor  4, 
and  1  remaining.  Hence  900  contains,  according  to  ( 1 1 7 — 2),  200 
times  the  divisor  4,  and  100  remaining.  We  multiply  the  divisor  4  by 
200,  and  subtract  the  product  800  from  986,  leaving  186  of  the  dividend 

yet  to  be  divided. 
ANALYSIS.  2.  We  know  that  18,  the  number 

4)986(200  expressed    by    the    two     left-hand 

4x200  =  800        40  figures  of  the  undivided    dividend, 

contains  4  times  4,  and  2  remaining. 

-'-  "  "  Hence    18   tens,    or    180,   contains, 

4x40      =160  6  according  to  (117—2),  40  times  4, 


4x6        =24 


n  n     o  4.  p  2        ^^^  20  remaining.      We  multiply 
^       the  divisor  4  by  40,  and  subtract 
the  product  160  from  186,  leaving 


2  26  yet  to  be  divided. 


12 


DI  VISION, 


3.  We  know  that  26  contains  6  times  4,  and  2  remaining,  wliich  is 
less  than  the  divisor ;  hence  the  division  is  completed. 

4.  We  have  now  found  that  there  are  200  fours  in  800,  40  fours  in 
160^,  and  6  fours  in  26,  and  2  remaining ;  and  we  know  that  800  +  160 
+  26  =  986.  Hence  986  contains  (200  +  40  +  6)  or  246  fours,  and  2 
remaining.  The  remainder  is  placed  over  the  divisor  and  written  after 
the  quotient ;  thus,  246|. 

SHORT     AND     Ij  O  N  G     DIVISION      C  O  M  P  A.  It  E  D  , 

121.  Compare  carefully  the  following  forms  of  writing  the 
work  in  division : 


(1- 

rOEM  USED  roR 

Two  Bteps  in  the 

4 

) 

EXPLANATION. 

process  written. 

)  986  (  200 

(2.) 

LONG  DIVISION. 

One  step  written. 

4  )  986  (  246 

(3.) 

SHOBT  DIVISION. 

Entirely  mental. 

4)986 

4x200= 

800 

40 

8 

■"2461 

186 

6 

18 

4x   40  = 

160 

246 

16 

26 

26 

4x      6  = 

24 

24 

To  divide  any  number  by  any  given 


Explanation. — 1.  We  find  how  many 
times  the  divisor  is  contained  in  the 
fewest  of  the  left-hand  figures  of  the 
dividend  which  will  contain  it. 

59  is  contained  3  times  in  215,  with  a 
remainder  38,  hence,  according  to 
(115—1),  it  is  contained  300  times  in 
21500,  with  a  remainder  3800. 

2.  We  annex  the  figure  in  the  next 
lower  order  of  the  dividend  to  the 
remainder  of  the  previous  division,  and  divide  the  number  thus  found 
by  the  divisor.     2  tens  annexed  to  380  tens  make  382  tens. 


129.    Prob.  II.— To 

divisor. 

1.  Divide  21524  by  59, 

59)21524(364 

177 

382 

354 

284 

236 

48 


DIVISION.  13 

59  is  contained  6  times  in  383,  with  a  remainder  28  ;  hence,  according 
to  (1 15 — 1),  it  is  contained  60  times  in  3820,  with  a  remainder  280. 
3.  We  annex  the  next  lower  figure  and  proceed  as  before. 

137.  Prob.  III. — To  divide  by  using  the  factors  of 
the  divisor. 

Ex.  1.  Divide  315  by  35. 

5)315  Explanation.— 1.  The  divisor  35  =  7  fives. 

7  nvesyol  fives  ^-  ^^^^^"^^  *l^°  ^^^   ^^  ^'  ''^  ^"^  *^'* 

J        I — ^ 315  =  63>«5.    (138.) 

3.  The    63  fives   contain    9  times  7  fives; 

hence  315  contains  9  times  7  fives  or  9  times  35. 

Ex.  2.  Divide  359  by  24. 
2  1^  5^ 
3  twos  I  1  7  9  twos        and  1  remaining  =      1 

4  (3  twos)     I  5  9  (3  tivos)  and  2  twos  remaining   =      4 
Quotient,       1  4  and  3  (3  twos)  remaining   =   18 

True  remainder,  2  3 

Explanation.— 1.  The  divisor  24  =  4x3x2  =  4  (3  twos). 

2.  Dividing  359  by  2,  we  find  that  359=179  twos  and  1  unit  remaining. 

3.  Dividing  179  twos  by  3  twos,  we  find  that  179  twos  —  59  (3  twos)  and 
2  twos  remaining. 

4.  Dividing  59  (3  twoi)  by  4  (3  twos),  we  find  that  59  (3  two^  contain 
4  (3  twos)  14  times  and  3  (3  twos)  remaining. 

Hence  359,  which  is  equal  to  59  (3  twos)  and  2  twos  +  1,  contains 
4  (3  twoi),  or  24,  14  times,  and  3  (3  twos)  +  2  twos  +  1,  or  23,  remaining. 

142.  Prob.  IV. — To  divide  v^^hen  the  divisor  con- 
sists of  only  one  order  of  units. 

1.  Divide  8736  by  500.  Explanation.— 1.  We  divide 
^  ^  ft  7  I  ^  r  ^^^*  ^^  *^®  factor  100.  This  is 
^  Lz2A2^                                      done  by  cutting  off  36,  the  units 

1  7  and  236  remaining.        and  tens  at  the  right  of  the  divi- 
dend. 

2.  We  divide  the  quotient,  87  hundreds,  by  the  factor  5,  which  gives 
a  quotient  of  17  and  2  hundred  remaining,  which  added  to  36  gives  236, 
the  true  remainder. 


14  DIVISION, 

DEFINITIONS. 

144.  Division  is  the  process  of  finding  how  many  times 
one  number  is  contained  in  another. 

145.  The  Dividend  is  the  number  divided. 

146.  The  Divisor  is  the  number  by  which  the  dividend 
is  divided. 

147.  The  Quotient  is  the  result  obtained  by  division. 

148.  The  Remainder  is  the  part  of  the  dividend  left 
after  the  division  is  performed. 

149.  A  Partial  Dividend  is  any  part  of  the  dividend 
which  is  divided  in  one  operation. 

150.  A  Partial  Quotient  is  any  part  of  the  quotient 
which  expresses  the  number  of  times  the  divisor  is  contained 
in  a  partial  dividend. 

151.  The  Process  of  Division  consists,  first,  in  find- 
ing the  partial  quotients  by  means  of  memorized  results ; 
second,  in  multiplying  the  divisor  by  the  partial  quotients  to 
find  the  partial  dividends ;  tMrd,  in  subtracting  the  partial 
dividends  from  the  part  of  the  dividend  that  remains  un- 
divided to  find  the  part  yet  to  be  divided. 

153.  Short  Division  is  that  form  of  division  in  which 
no  step  of  the  process  is  written. 

153.  Long  Division  is  that  form  of  division  in  which 
the  third  step  of  the  process  is  written. 

154.  The  S  iff  ft  of  Division  is  -^,  and  is  read  divided 
by.  When  placed  between  two  numbers,  it  denotes  that  the 
number  before  it  is  to  be  divided  by  the  number  after  it; 
thus,  28  -h  7  is  read,  28  divided  by  7. 

Division  is  also  expressed  by  placing  the  dividend  above  the  divisor, 
Avitli  a  short  horizontal  line  between  them ;  thus,  ^^-  is  read,  35 
divided  by  5. 


DIVISION.  15 

155.  Prin^ciples. — /.  Tlie  dividend  and  divisor  must 
he  nuinbers  of  the  same  denomination. 

II.  The  denomination  of  the  quotient  is  determined 
"by  the  nature  of  the  problem  solved. 

III.  The  remainder  is  of  the  same  denomination  as 
the  dividend. 

REVIEW   AND  TEST  QUESTIONS. 

156.  1.  Define  Division,  and  illustrate  each  step  in  the  process  by- 
examples. 

2.  Explain  and  illustrate    by  examples   Partial    Dividend,  Partial 
Quotient,  and  Remainder. 

3.  Prepare  two  examples  illustrating  each  of  the  f ollowhig  problems : 

I.  Given  all  the  parts,  to  find  the  whole. 
II.  Given  the  whole  and  one  of  the  parts,  to  find  the  other 

part. 
III.  Given  one  of  the  equal  parts  and  the  number  of  parts,  to 

find  the  whole. 
rV.  Given  the  whole  and  the  size  of  one  of  the  parts,  to  find 

the  number  of  parts. 
V.  Given  the  whole  and  the  number  of  equal  parts,  to  find 
the  size  of  one  of  the  parts. 

4.  Show  that  45  can  be  expressed  as  nines,  as  jives,  as  threes. 

5.  What  is  meant  by  true  remainder,  and  how  found  ? 

6.  Explain  division  by  factors.     Illustrate  by  an  example. 

7.  Why  cut  off  as  many  figures  at  the  right  of  the  dividend  as  there 
are  ciphers  at  the  right  of  the  divisor  ?    Illustrate  by  an  example. 

8.  Give  a  rule  for  dividing  by  a  number  with  one  or  more  ciphers  at 
the  right.     Illastrate  the  steps  in  the  process  by  an  example. 

9.  Explain  the  difference  between  Long  and  Short  Division,  and  show 
that  the  process  in  both  cases  is  performed  mentally. 

10.  Illustrate  each  of  the  following  problems  by  three  examples : 

VI.  Given  the  final  quotient  of  a  continued  division,  the  true 

remainder,  and  the  several  divisors,  to  find  the  dividend. 

VII.  Given  the  product  of  a  continued  multiplication  and  the 

several  multipliers,  to  find  the  multiplicand. 
VIII.  Given  the  sum  and  the  difference  of  two  numbers,  to  find 
the  numbers. 


X6  PROPERTIES     OF    NUMBERS. 

PEOPEETIES    OF    NUMBERS. 

163.  An  Integer  is  a  number  that  expresses  how  many 
there  are  in  a  collection  of  wJioIe  things. 

Thus,  8  yards,  12  Louses,  32  dollars. 

164.  An  Exact  JDivisor  is  a  number  that  will  divide 
another  number  without  a  remainder. 

Thus,  3  or  5  is  an  exact  divisor  of  15. 

All  numbers  with  reference  to  exact  divisors  are  either  prime  or 
composite. 

165.  A  JPrime  Number  is  a  number  that  has  no  exact 
divisor  besides  1  and  itself. 

Thus,  1,  3,  5,  7, 11,  13,  etc.,  are  prime  numbers. 

166.  A   Composite  JVtimber  is  a  number  that  has 
other  exact  divisors  besides  1  and  itself. 

Thus,  6  is  divisible  by  either  2  or  3  ;  hence  is  composite. 

167.  A  I^rime  I>ivisor  is  a  prime  number  used  as  a 
divisor. 

Thus,  in  35  -^-  7,  7  is  a  prime  divisor. 

168.  A  Composite  Divisor  is  a  composite  number 
used  as  a  divisor. 

Thus,  in  18  -i-  6,  6  is  a  composite  divisor. 


EXACT    DIVISION". 

169.  The  following  tests  of  exact  division  should  be  care- 
fully studied  and  fixed  in  the  memory  for  future  use. 

Piiop.  I. — A  divisor  of  any  number  is  a  divisor  ofanynum- 
her  of  times  that  numher. 

Thus,  12  =  3  fours.  Hence,  12  x  6  =  3  fours  x  6  =  18  fours.  But  18 
fours  are  divisible  by  4.    Hence,  12  x  6,  or  72,  is  divisible  by  4. 


EXACT    DIVISION.  17 

Prop.  II.— ^  divisor  of  each  of  two  or  more  numhers  is  a 
divisor  of  their  simi. 

Thus,  5  is  a  divisor  of  10  and  30 ;  that  is,  10  =  2  fives  and  30  =  6  fives. 
Hence,  10  +  30  =  2  fives  +  6  fives  =  8  fives.  But  8  fives  are  divisible  by  5. 
Hence,  5  Ls  a  divisor  of  the  sum  of  10  and  30. 

Prop.  III. — A  divisor  of  each  of  two  numiers  is  a  divisor  of 
their  differetice. 

Thus,  3  is  a  divisor  of  ^7  and  15 ;  that  is,  27  =  9  threes  and  15  = 
5  threes.  Hence,  27—15  =  9  threes  —  5  threes  =  4  threes.  But  4  threes 
are  divisible  hj  3.  Hence  3  is  a  divisor  of  the  difference  between  27 
and  15. 

Prop.  IV. — A7iy  numher  ending  with  a  cipher  is  divisible 
hy  the  divisors  of  10,  viz,,  2  and  5. 

Thus,  370  =  37  times  10.  Hence  is  divisible  by  2  and  5,  the  divisors 
of  10,  according  to  Prop.  I. 

Prop.  V. — A7iy  ^lumber  is  divisible  by  either  of  the  divisors 
of  10,  when  its  right-hand  figure  is  divisible  by  the  same. 

Thus,  498  =  490  +  8.  Each  of  these  parts  is  divisible  by  2.  Hence 
the  number  498  is  divisible  by  2,  according  to  Prop.  II, 

In  the  same  way  it  may  be  shown  that  495  is  divisible  by  5. 

Prop.  VI. — Any  number  endi^ig  with  two  ciphers  is  divisible 
by  the  divisors  of  100,  viz.,  2,  4,  5,  10,  20,  25,  and  50. 

Thus,  8900  —  89  times  100.  Hence  is  divisible  by  any  of  the  divisors 
of  100,  according  to  Prop.  I. 

Prop.  VII. — Any  number  is  divisible  by  any  one  of  the 
divisors  of  100,  when  the  number  expressed  by  its  two  right- 
hand  figures  is  divisible  by  the  same. 

Thus,  4975  =  4900  +  75.  Any  divisor  of  100  is  a  divisor  of  4900 
(Prop,  VI).  Hence,  any  divisor  of  100  which  will  divide  75  is  a  divisor 
of  4975  (Prop.  II). 

Prop.  VIII. — Any  numher  ending  with  three  ciphers  is 
divisible  by  the  divisors  of  1000,  viz.,  2,  4,  5,  8,  10,  20,  25,  40, 
50,  100,  125,  200,  250,  and  500. 

Thus,  83000  =  83  times  1000.  Hence  is  divisible  by  any  of  the  divi- 
Bors  of  1000,  according  to  Prop.  I. 


18  EXACT     DIVISION. 

Prop,  IX. — Any  number  is  divisible  by  any  one  of  the  divi- 
sors of  1000,  when  the  number  expressed  by  its  three  right-hand 
figures  is  divisible  by  the  same. 

Thus,  92625  =  92000  +  625.  Any  divisor  of  1000  is  a  divisor  of  92000 
(Prop.  VIII).  Hence,  any  divisor  of  1000  which  will  divide  625  is  a  divi- 
sor of  92625  (Prop.  II). 

Prop.  X. — Any  number  is  divisible,  by  9,  if  the  sum  of  its 
digits  is  divisible  by  9. 

This  proposition  may  be  shown  thus  : 

(1.)  486  =  400  +  80  +  6. 

(2.)  100  =  99  +  1  =:  11  nines  +  1.  Hence,  400  =  44  nines  +  4,  and  is 
divisible  by  9  with  a  remainder  4. 

(3.)  10  =  9  + 1  r=  1  nine  +  l.  Hence,  80  =  8  nines +  8,  and  is  divisible 
by  9  with  a  remainder  8. 

(4.)  From  the  foregoing  it  follows  that  400  +  80  +  6,  or  486,  is  divisible 
by  9  with  a  remainder  4  +  8  +  6,  the  sum  of  the  digits.  Hence,  if  the 
Bum  of  the  digits  is  divisible  by  9,  the  number  486  is  divisible  by  9 
(Prop.  II). 

Prop.  XI. — Any  number  is  divisible  by  3,  if  the  sum  of  its 
digits  is  divisible  by  3. 

This  proposition  is  shown  in  the  same  manner  as  Prop.  X  ;  as  3  divides 
10,  100, 1000,  etc.,  with  a  remainder  1  in  each  case. 

Prop.  XII. — A7iy  number  is  divisible  by  11,  if  the  difference 
of  the  sums  of  the  digits  in  the  odd  and  even  places  is  zero  or  is 
divisible  by  11. 

This  may  be  shown  thus : 

(1.)  4928  =  4000  +  900  +  20  +  8. 

(2.)  1000  =  91  elevens  —  1.    Hence,  4000  =  364  elevens  —  4. 

(3.)  100  =  9  elevens  +  1.    Hence,  900  =  81  elevens  +  9. 

(4.)  10  =  1  eleven  —  1.     Hence,  20  =  2  elevens  —  2. 

(5.)  From  the  foregoing  it  follows  that  4928  =  364  elevens  +  81  elevens 
+  2  elevens— 4  +  9-2  +  8. 

But  -  4  +  9  -  2  +  8  =  11.  Hence,  4928  =  364  elevens  +  81  elevens 
+  2  elevens +  1  eleven  =  448  elevens,  and  is  therefore  divisible  by  11. 

The  same  course  of  reasoning  applies  where  the  difference  is  minus 
or  zero. 


PRIME    NUMBERS.  J.9 

PRIME    NUMBEES. 

PRErAltATORY      fMOrOSITlONS. 

171.  Prop.  I. — All  even  numbers  are  divisible  by  2,  and 
consequently  all  even  mimbers,  exce]^t  2,  are  composite. 

Hence,  in  finding  the  prime  numbers,  we  cancel  as  composite  all  even 
numbers  except  2. 

Thus,  3,  4,  5,  0,  7,  $,  9,  X^,  11,  It,  and  so  on. 

Prop.  II. — Each  number  i7i  the  series  of  odd  numbers  is  2 
greater  than  the  number  immediately  ^preceding  it. 

Thus,  the  numbers  left  after  cancelling  the  even  numbers  are 

3  5  7  9  11  13,    and  so  on. 

3        3  +  2        5  +  2        7  +  2        9  +  2        11  +  2 
Prop.  III. — In  the  series  of  odd  numbers,   every  third 
number  from  3  is  divisible  by  3,  every  fifth  number  from  5  is 
divisible  by  5,  and  so  on  ivith  each  number  in  the  series. 

This  proposition  may  be  shown  thus  : 

According  to  Prop.  II,  the  series  of  odd  numbers  increase  by  2's. 
Hence  the  third  number  from  3  is  found  by  adding  2  three  times,  thus  : 

3  5  7  9 


3  3+2  3+2+2  3+2+2+2 

From  this  it  will  be  seen  that  9,  the  third  number  from  3,  is  composed 

of  3,  plus  3  twos,  and  is  divisible  by  3  (Prop.  II) ;  and  so  with  the  third 

number  from  9,  and  so  on. 

By  the  same  course  of  reasoning,  each  fifth  number  in  the  series, 

counting  from  5,  may  be  shown  to  be  divisible  by  5  ;  and  so  with  any 

other  number  in  the  series  ;  hence  the  following  method  of  finding  the 

prime  numbers. 

ILLZrSTIiATION      OF     P  It  O  C  E  S  S  . 

172.    Prob.— To  find  all  the  Prime  Numbers  from  1 
to  any  given  number. 

Find  all  the  prime  numbers  from  1  to  63. 


30 


FA  CTORING, 

1 

3 

5 

7            9 

3 

11 

13 

15 

17 

19 

21          23 

25 

27 

3    6 

3     1 

6 

3     9 

29 

31 

33 

35          37 

39 

41 

3     11 

5     1 

3     13 

43 

45 

47 

49           51 

53 

55 

3     5     9     16 

7               3     17 

6    11 

57  59  61  63 

3     19  3     7     9     31 

Explanation. — 1.  Arrange  tlie  series  of  odd  numbers  in  lines,  at  con- 
venient distances  from  each  otlier,  as  sliown  in  illustration. 

3.  Write  3  under  every  third  number  from  3,  5  under  ewevj  fifth  num- 
ber from  5,  7  under  every  seventh  number  from  7,  and  so  on  with  each 
of  the  other  numbers. 

3.  The  terms  under  which  the  numbers  are  written  are  composite, 
and  the  numbers  written  under  are  their  factors,  according  to  Prop.  III. 
All  the  remaining  numbers  are  prime. 

Hence  all  the  prime  numbers  from  1  to  63  are  1,  2,  3,  5,  7,  11,  13, 17, 
19,  23,  39,  31,  37,  41,  43,  47,  53,  59,  61. 


FAOTOEIISTG, 

175.  A  Factor  is  one  of  the  equal  parts  of  a  number,  or 
one  of  its  exact  divisors. 

Thus,  15  is  composed  of  3  fives  or  5  threes ;  hence,  5  and  3  are  factors 
of  15. 

176.  A  JPrifne  Factor  is  a  prime  number  which  is  a 
factor  of  a  given  number. 

Thus,  5  is  a  prime  factor  of  30. 

177.  A   Co^nposite  Factor  is  a  composite  number 
which  is  a  factor  of  a  given  number. 

Thus,  6  is  a  composite  factor  of  34. 

178.  Factoring  is  the  process  of  resolving  a  composite 
number  into  its  factors. 


CANCELLATION.  21 

1*79.  An  JExjwnent  is  a  small  figure  placed  at  the  right 
of  a  number  and  a  little  above,  to  show  how  many  times  the 
number  is  used  as  a  factor. 

Thus,  3^  =  3x3x3x3x3.  The  5  at  the  right  of  3  denotes  that  the  3 
is  used  5  times  as  a  factor. 

180.  A  Cofunion  Factor  is  a  number  that  is  a  factor 
of  each  of  two  or  more  numbers. 

Thus,  3  is  a  factor  of  6,  9,  12,  and  15 ;  hence  is  a  common  factor. 

181.  The  Greatest  Common  Factor  is  the  greatest 
number  that  is  a  factor  of  each  of  two  or  more  numbers. 

Thus,  4  is  the  greatest  number  that  is  a  factor  of  8  and  also  of  12. 
Hence  4  is  the  greatest  common  factor  of  8  and  12. 

IIjZUSTItA  TIO  N      OF     P  B  O  C  E  S  S . 

182.  Find  the  prime  factors  of  462. 

2)462  Explanation.— 1.  We  observe  that  the  number 

o  \~9V7  462  is  divisible   by  2,  the  smallest  prime  number. 

I — ! —  Hence  we  divide  by  2. 

7)77  2.  We  observe  that  the  first  quotient,  231,  is  divis- 

-,  1  ible  by  3,  which  is  a  prime  number.     Hence  we  divide 

by  3. 
8.  We  observe  that  the  second  quotient,  77,  is  divisible  by  7,  which  is 
a  prime  number.     Hence  we  divide  by  7. 

4.  The  third  quotient,  11,  is  a  prime  number.  Hence  the  prime  fac- 
tors of  462  are  2,  3,  7,  and  11 ;  that  is,  462  =  2  x  3  x  7  x  11. 


OAE^OELLATION. 

PB  E  PAIt  ATORT     P  R  O  P  O  S  I TJ  O  N  8, 

185.  Prop.  I. — Rejecting  a  factor  from  a  number  divides 
the  number  by  that  factor. 

Thus,  72  =  24  x  3.  Hence,  rejecting  the  factor  3  from  72,  we  have  24, 
the  quotient  of  72  divided  by  3. 

Prop.  II. — Dividing  both  dividend  and  divisor  by  the  same 
number  does  not  change  the  quotient. 


22  GREATEST     COMMON    DIVISOR. 

Thus,  60^12  =  20  threes -f-  4  threes  ==  5. 

Observe  that  the  unit  three,  in  20  threes  -s-  4  threes,  does  not  in  any 
way  affect  the  size  of  the  quotient ;  therefore,  it  may  be  rejected  and 
the  quotient  will  not  be  changed. 

Hence,  dividing  both  the  dividend  60  and  the  divisor  12  by  3  does 
not  change  the  quotient. 

ILLUSTRATION      OF     P H O C E 8 8 , 

186.    Ex.  1.  Divide  462  by  42. 

6  )  462       77  Explanation.— We  divide  both  the 

n  \  ~^  ^^^  ~w"  ^^  -'■-'■•  divisor  and  dividend  by  6.     According  to 

Prop.  II,  the  quotient  is  not  changed. 
Hence,  77-^7  =  462-^42  =  11. 

Ex.  2.  Divide  65  x  24  x  55  by  39  x  15  x  35. 

•Hi' 


Explanation.— 1.  We  divide  any  factor  in  the  dividend  by  any  num- 
ber that  will  divide  a  factor  in  the  divisor. 

Thus,  65  in  the  dividend  and  15  in  the  divisor  are  divided  each  by  5. 
In  the  same  manner,  55  and  35, 13  and  39,  24  and  3  are  divided. 

The  remaining  factors,  8  and  11,  in  the  dividend  are  prime  to  each  of 
the  remaining  factors  in  the  divisor.  Hence,  no  further  division  can  be 
performed. 

2.  We  divide  the  product  of  8  and  11,  the  remaining  factors  in  the 
dividend,  by  the  product  of  3  and  7,  the  remaining  factors  in  the  divisor, 
and  find  as  a  quotient  4/y>  which,  according  to  (185 — II),  is  equal  to 
the  quotient  of  65  x  24  x  55  divided  by  39  x  15  x  35. 

All  similar  cases  may  be  treated  in  the  same  manner. 


GREATEST    OOMMOE    DIVISOR. 

190.  A   Common  Divisor  is  a  number  that  is  an 
exact  divisor  of  each  of  two  or  more  numbers. 
Thus,  5  is  a  divisor  of  10, 15,  and  20. 


1.^            8            11 

8  X  11 

88 

$0  X  10  X  $$ 
3^7 

3x7 

~  21 

GREATEST     COMMON    DIVTSOR.  23 

191.  The  Greatest  Connnon  Divisor  is  the  great- 
est number  that  is  an  exact  divisor  of  each  of  two  or  more 
numbers. 

Thus,  3  is  the  greatest  exact  divisor  of  each  of  the  numbers  6  and  15. 
Hence  3  is  their  greatest  common  divisor. 

192.  Numbers  are  prime  to  each  other  when  they 
have  no  common  divisor  besides  1 ;  thus,  8,  9,  25. 

METHOD     BY    FACTOEING. 

ritEBAItATORT     PROPOSITION, 

193.  Illustrate  the  following  proposition  by  examples. 

Tlie  greatest  common  divisor  is  the  product  of  the  prime  faO' 
tors  that  are  common  to  all  the  given  numbers  ;  thus, 

42=    7x2x3=    7  sixes; 
66  =  11  X  2  X  3  =  11  sixes. 

7  and  11  being  prime  to  each  other,  6  must  be  the  greatest  common 
divisor  of  7  sixes  and  11  sixes.  But  6  is  the  product  of  2  and  3,  the  com- 
mon prime  factors ;  hence  the  greatest  common  divisor  of  42  and  64  is 
the  product  of  their  common  prime  factors. 

ILLUSTRATION     OF     PROCESS. 

194.  Prob.  I.— To  find  the  Greatest  Common  Divisor 
of  two  or  more  numbers  by  factoring". 

Find  the  greatest  common  divisor  of  98,  70,  and  154. 

(1.)  (2.) 

2)  98      2)^70      2)154                    2)98  70      154 

7)49      7)35         7)77        Or,       7)49  35         77 

"7               5               11                            7  5         11 

2x7  =  greatest  common  divisor. 


M  GREATEST     COMMON    DIVISOR. 

Explanation. — 1.  We  resolve  each  of  tlie  numbers  into  their  prime 
factors,  as  shown  in  (1)  or  (2). 

2.  We  observe  that  2  and  7  are  the  only  prime  factors  common  to  all 
the  numbers.  Hence  the  product  of  2  and  7,  or  14,  according  to  (193), 
is  the  greatest  common  divisor  of  98,  70,  and  1 54. 

The  greatest  common  divisor  of  any  two  or  more  numbers  is  found  in 
the  same  manner. 

METHOD    BY    DIVISION. 

mErAItATORT     rJiOrOSlTIONS. 

107.  Let  the  two  following  propositions  be  carefully  studied 
and  illustrated  by  other  examples,  before  attempting  to  find 
the  greatest  common  divisor  by  this  method. 

Prop.  I. —  The  greatest  commoii  divisor  of  two  numbers  is 
the  greatest  common  divisor  of  the  smaller  number  and  their 
difference. 

Thus,  3  is  the  greatest  common  divisor  of  15  and  37. 

Hence,  15  =  5  threes     and     27  =  9  threes  ; 
and    9  threes  —  5  threes  ==  4  threes. 

But  9  and  5  are  prime  to  each  other  ;  hence,  4  and  5  must  be  prime  to 
each  other,  for  if  not,  their  common  divisor  will  divide  their  sum,  accord- 
ing to  (169—11),  and  be  a  common  divisor  of  9  and  5. 

Therefore,  3  is  the  greatest  common  divisor  of  5  threes  and  4  threes, 
or  of  15  and  12.  Hence,  the  greatest  common  divisor  of  two  numbers  is 
the  greatest  common  divisor  of  the  smaller  number  and  their  difference. 

Pkop.  II. — The  greatest  common  divisor  of  two  numbers  is 
the  greatest  common  divisor  of  the  smaller  number  and  the 
rejnainder  after  the  division  of  the  greater  by  the  less. 

This  proposition  may  be  illustrated  thus : 

22  —  6  =  16  1-  Subtract  6  from  22,  then  from  the  differ- 

■j^g  g  __  ^Q  ence,  16,  etc.,  until  a  remainder  less  than  6  is 

^^        r,  .  obtained. 

2.  Observe  that  the  number  of  times  6  has 
been  subtracted  is  the  quotient  of  22  divided  by  6,  and  hence  that  the 
remainder,  4,  is  the  remainder  after  the  division  of  22  by  6. 

3.  According  to  Prop.  1,  the  greatest  common  divisor  of  22  and  G  is 


LEAST     COMMON     MULTIPLE.  25 

the  greatest  common  divisor  of  tlieir  diflference,  16,  and  6.  It  is  also, 
according  to  the  same  Proposition,  the  greatest  common  divisor  of  10 
and  6,  and  of  4  and  6.  But  4  is  the  remainder  after  division  and  6  the 
smaller  number.  Hence  the  greatest  common  divisor  of  22  and  G  is  the 
greatest  common  divisor  of  the  smaller  numher  and  the  remainder  after 
division. 

ILLVSTRJiTION      OF     1*  It  O  C  i:  S  S  , 

198.   Pkob.  II. — To  find  the  Greatest  Common  Divisor 
of  two  or  more  numbers  by  continued  division. 

Find  the  greatest  common  divisor  of  28  and  176. 

2  8)170(6  Explanation.— 1.  We  divide  176 

1  g  Q  by  28  and  find  8  for  a  remainder ;  then 

we  divide  28  by  8,  and  find  4  for  a  re- 

o  )  /i  8  (  o  mainder  ;  then  we  divide  8  by  4,  and 

2  4  find  0  for  a  remainder. 

2  \  «  /"  2  ^*  -According  to  Prop.  II,  the  great- 

'  ^  est  common  divisor  of  28  and  176  is  the 

_  same  as  the  greatest  common  divisor  of 

0  28  and  8,  also  of  8  and  4.    But  4  is  the 

greatest  common   divisor  of  8  and  4. 

Hence  4  is  the  greatest  common  divisor  of  28  and  176. 


LEAST    COMMON    MULTIPLE. 

rnEJ^AItATOltY      PROPOSITIONS. 

211.  Prop.  I. — A  multiple  of  a  numher  contains  as  a  factor 
each  prime  factor  of  the  number  as  many  times  as  it  eiiters  into 
the  numher. 

Thus,  60,  which  is  a  multiple  of  12,  contains  5  times  12,  or  5  times 
2x2x3,  the  prime  factors  of  12.  Hence,  each  of  the  prime  factors  of 
12  enters  as  a  factor  into  60  as  many  times  as  it  enters  into  12. 

Prop.  II. — Tlie  least  common  multiple  of  ttco  or  more  given 
numhers  must  contain,  as  a  factor,  each  prime  factor  in  those 
numhers  the  greatest  numher  of  times  that  it  enters  into  any 
one  of  them. 

2 


26  LEAST     COMMON    MULTIPLE, 

Thus,  12  =  2  X  2  X  3,  and  9=3x3.  The  prime  factors  in  12  and  9  are 
2  and  3.  A  multiple  of  12,  according  to  Prop.  I,  must  contain  2  as  a 
factor  twice  and  3  once.  A  multiple  of  9,  according  to  the  same  propo- 
sition, must  contain  3  as  a  factor  twice.  Hence  a  number  which  is  a 
multiple  of  both  12  and  9  must  contain  2  as  a  factor  twice  and  3  twice, 
which  is  equal  to  2  x  2  x  3  x  3  =  36.  Hence  36  is  the  least  common  mul- 
tiple of  12  and  9. 

DEFINITIONS. 

212.  A  Multiple  of  a  number  is  a  number  that  is 
exactly  divisible  by  the  given  number. 

Thus,  24  is  divisible  by  8  ;  hence,  24  is  a  multiple  of  8. 

213.  A  Common  Multifile  of  two  or  more  numbers 
is  a  number  that  is  exactly  divisible  by  each  of  them. 

Thus,  36  is  divisible  by  each  of  the  numbers  4,  9,  and  12 ;  hence,  36  . 
is  a  common  multiple  of  4,  9,  and  12. 

214.  The  Least  Common  Multiple  of  two  or  more 

numbers  is  the  least  number  that  is  exactly  divisible  by  each 
of  them. 

Thus,  24  is  the  least  number  that  is  divisible  by  each  of  the  numbers 
6  and  8 ;  hence,  24  is  the  least  common  multiple  of  6  and  8. 

METHOD    BY    FACTORING. 

ILLUSTRATION      OF      1*  R  O  C  E  S  S . 

215.  Prob.  I. — To  find,  by  factoring,  the  least  com- 
mon multiple  of  two  or  more  numbers. 

Find  the  least  common  multiple  of  18,  24,  15,  and  35. 

Explanation. — 1.  We  observe 
that  3  is  a  factor  of  18,  24,  and  15. 
Dividing  these  numbers  by  3,  wo 
write  the  quotients  with  35,  in  the 
second  line. 

2.  Observing  that  2  is  a  factor  of  G 
and  8,  we  divide  as  before,  and  fin*'  the  third  line  of  numbers.     Dividing 


3 

18 

24 

15 

35 

2 

G 

8 

5 

35 

5 

3 

4 

5 

35 

3 

4 

7 

LEAST     C03IM0X    MULTIPLE.  2tt 

by  5,  we  find  the  fourth  line  of  numbers,  which  are  prime  to  each 
other  ;  hence  cannot  be  further  divided. 

3.  Observe  the  divisors  3,  2,  and  5  are  all  the  factors  that  are  common 
to  any  two  or  more  of  the  given  numbers,  and  the  quotients  3,  4,  and  7 
are  the  factors  that  belong  each  only  to  one  number.  Therefore  the 
divisors  and  quotients  together  contain  each  of  the  prime  factors  of  18, 
24,  15,  and  35  as  many  times  as  it  enters  into  any  one  of  these  numbers. 
Thus,  the  divisors  3  and  2,  with  the  quotient  3,  are  the  i)rime  factors  of 
18  ;  and  so  with  the  other  numbers. 

Hence,  according  to  (211 — II),  the  continued  product  of  the  divisors 
3,  2,  and  5,  and  the  quotients  3,  4,  and  7,  which  is  equal  to  2520,  is  the 
least  common  multiple  of  18,  24,  15,  and  35. 


METHOD    BY    GKEATEST    COMMON    DIVISOR. 

ILLUSTRATION      OF     PROCESS, 

218.  Pkob.  1 1. — To  find,  by  using  the  greatest  com- 
mon divisor,  the  least  common  multiple  of  two  or  more 
numbers. 

Find  the  least  common  multiple  of  195  and  255. 
Explanation. — 1.  We  find  the  greatest  common  divisor  of  195  and 
255,  wliicli  is  15. 

2.  The  greatest  common  divisor,  15,  according  to  (19J5),  contains  all 
tlie  prime  factors  that  are  common  to  195  and  255.  Dividing  each  of 
tliese  numbers  by  15,  we  find  tlie  factors  that  are  not  common,  namely, 
13  and  17. 

3.  The  common  divisor  15  and  the  quotient  13  contain  all  the  prime 
factors  of  195,  and  the  common  divisor  15  and  the  quotient  17  contain  all 
the  prime  factors  of  255. 

Hence,  according  to  (211 — II),  the  continued  product  of  the  common 
divisor  15  and  the  quotients  13  and  17,  which  is  3315,  is  the  least  com- 
mon multiple  of  195  and  255. 

The  least  common  multiple  of  any  two  numbers  is  found  in 
the  same  manner ;  and  of  three  or  more  by  finding  first  the 
least  common  multiple  of  two  of  them,  then  finding  the  least 
common  multiple  of  the  multiple  thus  found  and  the  third 
number,  and  so  on. 


f8  PROPERTIES     OF    NUMBERS. 


REVIEW    AND    TEST    QUESTIONS. 

223.  1.  Define  Prime  Number,  Composite  Number,  and 
Exact  Divisor,  and  illustrate  each  by  an  example. 

2.  What  is  meant  by  an  Odd  Number  ?    An  Even  Number  ? 

3.  Show  that  if  an  even  number  is  divisible  by  an  odd  num- 
ber, the  quotient  must  be  even. 

4.  Name  the  prime  numbers  from  1  to  40. 

5.  Why  are  all  even  numbers  except  2  composite  ? 

G.  State  how  you  would  show,  in  the  series  of  odd  numbers, 
that  every  fifth  number  from  5  is  divisible  by  5. 

7.  What  is  a  Factor  ?    A  Prime  Factor  ? 

8.  What  are  the  prime  factors  of  81  ?    Of  64  ?    Of  125  ? 

9.  Show  that  rejecting  the  same  factor  from  the  divisor  and 
dividend  does  not  change  the  quotient. 

10.  Explain  Cancellation,  and  illustrate  by  an  example. 

11.  Give  reasons  for  calling  an  exact  divisor  a  measure. 

12.  What  is  a  Common  Measure  ?  The  Greatest  Common 
Measure  ?    Illustrate  each  answer  by  an  example. 

13.  Show  that  the  greatest  common  divisor  of  42  and  114 
is  the  greatest  common  divisor  of  42  and  the  remainder  after 
the  division  of  114  by  42. 

14.  Explain  the  rule  for  finding  the  greatest  common 
divisor  by  factoring ;  by  division. 

15.  Why  must  we  finally  get  a  common  divisor  if  the  greater 
of  two  numbers  be  divided  by  the  less,  and  the  divisor  by  the 
remainder,  and  so  on  ? 

16.  What  is  a  Multiple  ?    The  Least  Common  Multiple  ? 

17.  Explain  how  the  Least  Common  Multiple  of  two  or 
more  numbers  is  found  by  using  their  greatest  common  divisor. 

18.  Prove  that  a  number  is  divisible  by  9  when  the  sum  of 
its  digits  is  divisible  by  9. 

19.  Prove  that  a  number  is  divisible  by  11  when  the  differ- 
ence of  the  sums  of  the  digits  in  the  odd  and  even  places  is  zero. 


REDUCTION     OF    FRACTIONS,  29 


FEACTIONS. 

225.  A  Fractional  Unit  is  one  of  the  equal  parts  of 
anything  regarded  as  a  whole. 

226.  A  Fraction  is  one  or  more  of  the  equal  parts  of 
anything  regarded  as  a  whole. 

227.  The  Unit  of  a  Fraction  is  the  unit  or  whole 
which  is  considered  as  divided  into  equal  parts. 

228.  The  Numerator  is  the  number  above  the  dividing 
line  in  the  expression  of  a  fraction,  and  indicates  how  many 
equal  parts  are  in  the  fraction. 

229.  The  Denominator  is  the  number  below  the 
dividing  line  in  the  expression  of  a  fraction,  and  indicates 
how  many  eq^ial  parts  are  in  the  whole. 

230.  The  Terms  of  a  fraction  are  the  numerator  and 
denominator. 

231.  Taken  together,  the  terms  of  a  fraction  are  called  a 
Fraction^  or  Fractional  I^iunher, 

232.  Hence,  the  word  Fraction  means  one  or  more  of 
the  equal  parts  of  anything,  or  the  expression  that  denotes  one 
or  more  of  the  equal  parts  of  anything. 

PRINCIPLES    OF    REDUCTION 
235.  Illustrate  the  following  principles  with  examples : 

Piii;n".  I. — Tlie  numerator  and  denominator  of  a  frac- 
tion  represent,  each,  parts  of  the  same  size ;  thus, 
3  4 

7  5 

(1.)    ^^^^ (2.) 


Observe  in  illustration  (1)  the  denominator  7  represents  the  whole  or 
7  sevenths,  and  the  numerator  3  represents  3  sevenths  ;  in  illustration  (2), 
the  denominator  represents  ^fifths,  and  the  numerator  Affihs.  Hence  the 
numerator  and  denominator  of  a  fraction  represent  parts  of  the  same  size. 


30  REDUCTION     OF    FRACTIONS, 

Prin",  IL — Multiplying  both  the  terms  of  a  fraction 
hy  the  same  number  does  not  change  the  value  of  the 
fraction;  thus, 

2  _  ?  X  4_^ 

3  ""  3  X  4  "■  12 


Be  particular  to  observe  in  the  illustration  that  the  amount  expressed 
by  the  2  in  the  numerator  or  the  3  in  the  denommator  of  |  is  not  changed 
by  making  each  part  into  4  equal  parts  ;  therefore  f  and  -f^  express,  each, 
the  same  amount  of  the  same  whole. 

Hence,  multiplying  the  numerator  and  denominator  by  the  same  num- 
ber means,  so  far  as  the  real  fraction  is  concerned,  dividing  the  equal 
parts  in  each  into  as  many  equal  parts  as  there  are  units  in  the  nmnber  hy 
which  they  are  multiplied. 

Prin.  hi. — Dividing  both  terms  of  a  fraction  by  the 
same  nurnber  does  not  change  the  value  of  the  fraction  ; 
thus,  ^  -^  3  _  3 

12  -^  3  ~  4 


The  amount  expressed  by  the  9  in  the  numerator  or  the  12  m  the  de- 
nominator of  /^  is  not  changed  by  putting  every  3  parts  into  one,  as  will 
be  seen  from  the  illustration. 

Hence  -^^  and  f  express  each  the  same  amount  of  the  same  whole,  and 
dividing  the  numerator  and  denominator  by  the  same  number  means 
putting  as  many  parts  in  each  into  one  as  there  are  units  in  the  number  by 
which  they  are  divided. 

DEFmiTIONS. 

336.  The  Value  of  a  fraction  is  the  amount  which  it 
represents. 

237.  deduction  is  the  process  of  changing  the  terms  of 
a  fraction  without  altering  its  vakie. 

238.  A  fraction  is  reduced  to  Higher  Terms  when  its 
numerator  and  denominator  are  expressed  by  larger  numbers. 
Thus,  i  =  A- 


REDUCTION     OF    FB  ACTIONS.  31 

239.  A  fraction  is  reduced  to  Lower  Terms  when  its 
numerator  and  denominator  are  expressed  by  smaller  numbers. 
Thus,  A  =  f 

240.  A  fraction  is  expressed  in  its  Loivest  Terms  when 
its  numerator  and  denominator  are  2^ rime  to  each  other. 

Thus,  in  I,  the  numerator  and  denominator  4  and  9  are  prime  to  each 
other ;  hence  the  fraction  is  expressed  in  its  lowest  terms. 

241.  A  Common  Denominator  is  a  denominator 
that  belongs  to  two  or  more  fractions. 

242.  The  Least  Cominon  I^enominator  of  two 

or  more  fractions  is  the  least  denominator  to  which  they  can 
all  be  reduced. 

243.  A  ^roxier  Ft^action  is  one  whose  numerator  is 
less  than  the  denominator ;  as  f ,  ^. 

244.  An  Imiiroper  Fraction  is  one  whose  numerator 
is  equal  to,  or  greater  than,  the  denominator  ;  as  |,  |. 

245.  A  Mixed  dumber  is  a  number  composed  of  an 
integer  and  a  fraction ;  as  4f ,  I84. 

ILl^USTJtATION     OF     P  It  O  C  E  S  8 , 

246.  Prob.  I. — To  reduce  a  whole  or  mixed  number 
to  an  improper  fraction. 

1.  Reduce  3f  equal  lines  to  fifths. 


=  IS  fifths  ^  ~ 


Explanation, — Each  whole  line  is  equal  to  5  fifths,  as  shown  in 
the  illustration;  3  lines  must  therefore  be  equal  to  15  fift?is.  15  fifths 
+  3  fifths  =  18  fifths.    Hence  in  3|  lines  there  are  Y  of  a  line. 


33  REDUCTION     OF    FRACTIONS. 

249.     Prob.  II. — To  reduce  an  improper  fraction  to 
an  integer  or  a  mixed  number. 

1.  Keduce  ^fourths  of  a  line  to  whole  lines. 

f  =  9  -^  4  =  2i 


Explanation. — A  whole  line  is  composed  of  4  fourths.  Hence,  to 
make  tlie  9  fourtJis  of  a  line  into  whole  lines,  we  put  every  four  parts 
into  one,  as  shown  in  the  illustration,  or  divide  the  9  by  4,  which  gives 
2  wholes  and  1  of  the  fourths  remaining. 

352.  Prob.  III. — To  reduce  a  fraction  to  higher 
terms. 

1.  Keduce  f  of  a  line  to  twelfths. 

2  __  2  X  4:_^ 

3  ~  3  X  4  ~  12 


Explanation. — 1.  To  make  a  whole,  which  is  already  in  thirds,  into 
12  equal  parts,  each  third  must  be  made  mto  four  equal  parts, 

2.  The  numerator  of  the  given  fraction  expresses  2  thirds,  and  the 
denominator  3  thirds  ;  making  each  third  in  both  into  four  equal  parts 
(ti45 — II),  as  shown  in  the  illustration,  the  new  numerator  and  denom- 
inator will  each  contain  4  times  as  many  parts  as  in  the  given  fraction. 

Hence,  |  of  a  line  is  reduced  to  twelfths  by  multiplying  both  numera- 
tor and  denominator  by  4. 

lower 


255. 

Pr 

OB. 

IV.— To 

reduce   a 

fraction    t< 

terms. 

Keduce 

A 

of  a 

given  line 

to  fourths. 

9 
12 

-^3 
^3 

= 

3 
4 

., 

.^ 

«■■ 

■       MMM        MM 

m                   

— M.      MM      ^M 

ADDITION     OF    FRACTIONS.  33 

Explanation. — 1.  To  make  into  4  ^qual  parts  or  fourths  a  whole, 
which  is  already  in  12  equal  parts,  or  twelfths,  every  3  of  the  12  parts 
must  be  put  into  one. 

2.  The  numerator  of  the  given  fraction  expresses  9  twelfths,  and  the 
denominator  12  twelfths;  putting  every  3  twelfths  into  one,  in  both 
(235 — III),  as  shown  in  the  illustration,  the  new  numerator  and 
denominator  will  each  contain  one  third  as  many  parts  as  in  the  given 
fraction. 

Hence,  ^-^  of  a  line  is  reduced  to  fourths  by  dividing  both  numerator 
and  denominator  by  3. 

258.  Prob.  V. — To  change  fractions  to  equivalent 
ones  having  a  common  denominator. 

1.  Reduce  |  and  f  of  a  line  to  fractions  haying  a  common 
denominator. 


(1-) 


(2.) 

Explanation.— 1.  We  find  the  least  common  multiple  of  the  denom- 
inators 3  and  4,  which  is  12. 

8.  We  reduce  each  of  the  fractions  to  twelfths  (252 — III),  as  shown 
in  illustrations  (1)  and  (2). 


ADDITION     AND    SUBTRACTION. 

PJtEP  AR  ATOMY    1*  B  O  P  O  S  I  T I  O  N  S  . 

261.  Prop.  I. — Fractional  units  of  the  same  kind,  that 
are  fractions  of  the  same  whole,  are  added  or  subtracted  in  the 
same  manner  as  integral  units. 

Thus,  I  of  a  yard  can  be  added  to  or  subtracted  from  f  of  a  yard, 
because  they  are  each  fifths  of  one  yard.  But  |  of  a  yard  cannot  be 
added  to  or  subtracted  from  |  of  a  day. 


2 

3       .           "" 

2x4        8 
3x4       12 

^■1^       B^^H 

■  HBIH       BiBHH 

■■^iH      H^HM      m^t^ 

3 

4                  ~ 

3x3        9 
4  X  3  ~  12 

^—     ■— ■     ^mm                                 m 

^^        ^^        ^^        ^^                                   «M_        M»M        __«        M«M 

4      32 

9~72 

5  60 

6  ""72 

155 

■  ~  72 

7       63 
8~72. 

34  SUBTRACTION     OF    FRACTIONS. 

Prop.  II. — Fractions  expressed  in  different  fractional  units 
rmist  he  changed  to  equivalent  fractions  having  the  same  frac- 
tional unit,  before  they  can  he  added  or  subtracted. 

263.  Prob.  I. — To  find  the  sum  of  any  two  or  more 
given  fractions. 

1.  Find  the  sum  of  f  +  |-  +  |. 

Explanation. — 1.  We  reduce  the 
fractions  to  tlie  satae  fractional  unit, 
by    reducing    them    to    their    least 
common   denominator,   which   is  72 
^  -^  =  m-        (258-V). 

2.  We  find  the  sum  of  the  numer- 
ators, 155,  and  write  it  over  the  com- 
mon denominator,  72,  and  reduce  VV- 
to  2H. 

The  sum  of  any  number  of  fractions  may  be  found  in  the 
same  manner. 

366.  Prob.  I.— To  find  the  diflferenee  of  any  two 
given  fractions. 

1.  Find  the  difference  between  f  and  -^. 

7        5        21       10       11  Explanation.  —  1.   We   reduce 

o        12  ^^  24        24  ^^^  24  *^®  given   fractions  to  their  least 

common  denominator,  which  is  24, 

2.  We  find  the  difference  of  the  numerators,  21  and  10,  and  write  it 
over  the  common  denominator,  giving  ^,  the  required  difference. 

2.  Find  the  difference  between  35 1  and  16|. 

35f  =  35-^^  Explanation.— 1.  We  reduce  the  |  and  f  to 

■^Q3  --  IQJL.  their  least  common  denominator. 

— ~  2.  ^\  cannot  be  taken  from  /^ ;  hence,  we  in- 

l^H"  crease  the  -^%  by  f|  or  1,  taken  from  the  35.    We 

now  subtract  -^^  from  ff ,  leaving  i|. 

3.  We  subtract  16  from  the  remaining  34,  leaving  18,  which  united 
with  1^  gives  18^-,  the  required  difference. 


3IULTTP  LIGATION     OF    FRACTIONS.  35 

MULTIPLICATION. 

PREPARATORY     PROPOSITIONS. 

269,  The  following  propositions  must  be  mastered  per- 
fectly, to  understand  and  explain  the  process  in  multiplica- 
tion and  division  of  fractions. 

Prop.  I. — Multiplying  the  numerator  of  a  fraction,  while 
the  denominator  remai^is  unchanged,  multiplies  the  fraction  ; 
thus, 

2  X  4  _  8 

5  ~"  5 


Observe  that  since  the  denominator  is  not  changed,  the  size  of  the 
parts  remain  the  same.  Hence  the  fraction  |  is  multiplied  by  4,  as 
shown  in  the  illustration,  by  multiplying  the  numerator  by  4. 

Peop.  IL — Dividing  the  denominator  of  a  fraction  ivhile 
the  numerator  remains  unchanged  multiplies  the  fraction  j 
thus, 

A  -  ? 

12  -^  4  ~  3 


(1.)    (2.) 


Observe  that  in  (1)  the  whole  is  made  into  1 2  equal  parts.  By  put- 
ting every  4  of  these  parts  into  one,  or  dividing  the  denominator  by  4, 
the  whole,  as  shown  in  (2),  is  made  into  3  equal  parts,  and  each  of  the 
2  parts  in  the  numerator  is  4  times  1  twelfth. 

Hence,  dividing  the  denominator  of  -^^  by  4,  the  number  of  parts  in 
the  numerator  remaining  the  same,  multiplies  the  fraction  by  4. 

Prop.  III. — Dividing  the  ^lumerator  of  a  fraction  while  the 
denominator  remains  unchanged  divides  the  fraction  ;  thus, 

6-^3  _  2 

9                      "9 
(l.)iof  (^^^^^ . 


36  MULTIPLICATION     OF    FRACTIONS, 

In  (1)  the  numerator  6  expresses  the  parts  taken,  and  one-third  of 
these  6  parts,  as  shown  by  comparing  (1)  and  (2),  the  denommator 
remaining  the  same,  is  one-third  of  the  value  of  the  fraction.  Hence, 
the  fraction  f  is  divided  by  3  by  dividing  the  numerator  by  3. 

Prop.  IV. — Multiplying  the  denominatoi-  of  a  fraction  luliile 
the  nvmerator  remains  unchanged  divides  the  fraction  ;  thus, 

5x2  ~  10 

(!•)  m (^0  I", 

In  (1)  the  whole  is  made  into  5  equal  parts  ;  multiplying  the  denom- 
inator by  2,  or  making  each  of  these  5  parts  into  2  equal  parts,  as  shown 
in  (2),  the  whole  is  made  into  10  equal  parts,  and  the  3  parts  in  the 
numerator  are  one-half  the  size  they  were  before. 

Hence,  multiplying  the  denominator  of  f  by  2,  the  numerator  remain- 
ing the  same,  divides  the  fraction  by  2. 

ILJLUSTHATION     OF     I*  It  O  C  E  S  S  . 

271.   Prob.  I. — To  multiply  a  fraction  by  an  integer. 

1.  Multiply  i  by  7. 

Solution. — 1.  According  to  (2G9 — I),  multiplying  the  numerator, 

the  denominator  remaining  the  same,  multiplies  the  fraction.     Hence,  7 

,  ,    4  X  7      28      „, 
times  f  IS  equal  to  -         =  —  =  3J. 
y  " 

2.  According  to  (269 — II),  a  fraction  is  also  multiplied  by  dividing 
the  denominator. 

274.    Prob.  II. — To  find  any  given  part  of  an  integer, 

or  To  multiply  an  integer  by  a  fraction, 

1.  Find  I  of  S395. 

Solution.— 1.  We  find  the  \  of  $395  by  dividing  it  by  5.  Hence  the 
first  step,  1395  -^  5  =  $79. 

2.  Since  |79  is  1  fifth  of  $395,  four  times  $79  will  be  4  fifths.  Hence 
the  second  step,  $79  x  4  =  $316. 

To  avoid  fractions  until  the  final  result,  we  multiply  by  the  numerator 
first,  then  divide  by  the  denominator. 


MULTIPLICATION     OF    FRACTIONS.  37 

276.    Prob.  III. — To  find  any  given  part  of  a  fraction, 

or  To  multiply  a  fraction  by  a  fraction. 

Find  the  f  of  j  of  a  given  line. 

FIKST  STEP. 

1     „  3  3  3 

of 


3  4  4  X  3       12 


SECOND  STEP. 

_3_  X  2  _  A  — i 

12  ~  12  ~  2 


Explanation. — According  to  (269 — IV),  a  fraction  is  divided  by 
multiplying  its  denominator.  Hence  we  find  the  \  of  f  by  multiplying 
the  denominator  4  by  3,  as  shown  in  First  Step. 

Having  found  1  third  of  f,  we  find  2  thirds,  according  to  (2G9 — I), 
by  multiplying  the  numerator  of  ^^  by  2,  as  shown  in  Second  Step. 

In  practice,  the  Second  Step  is  usually  made  the  First. 

379.   Pkob.  IV. — To  multiply  by  a  mixed  number. 

Multiply  372  by  64. 

372  Explanation.  —  1.  In    multiplying    by    a    mixed 

g  g  number,  the  multiplicand  is  taken  separately  (92)  as 

many  times  as  there  are  units  in  the  multiplier,  and 

'^  -^  *^ ''  such  a  part  of  a  time  as  is  indicated  by  the  fraction  in 

2  6  5  -^  the  multiplier ;  hence, 

249^5  2.  We  multiply  372  by  6|  by  multiplying  first  by  6, 

*  which  gives  the  product  2232,  and  adding  to  this  pro- 
duct f  of  372,  which  is  265 f  (2  74),  giving  2497f ,  the  product  of  372  and  6f . 

383.  Prob.  V.— To  multiply  when  both  multiplicand 
and  multiplier  are  mixed  numbers. 

Multiply  86|  by  54|. 

(1.)  s^=^^',      u\=^K 

(2.)    ^  X  ^  =  -a  Vi^^  =  47411-J. 


38  DIVISION     OF    FRACTIONS. 

Explanation. — 1.  We  reduce,  as  shown  in  (1),  both  multiplicand  and 
multiplier  to  improper  fractions. 

3.  We  multiply,  as  shown  in  (2),  the  numerators  together  for  the 
numerator  of  the  product,  and  the  denominators  together  for  the  denom- 
inators of  the  product  (275),  then  reduce  the  resulUto  a  mixed  number. 

DIVISION. 

IZLUSTRATION      OF     PROCESS. 

287.    Pkob.  I. — To  divide  a  fraction  by  an  integer. 

1.  Divide  f  by  4. 

8  8  -r-  4       2  Explanation.  —  1.  Accord- 

(-*  •)    9  "^  9  ~  9  ing  to  (269—111),  a  fraction  is 

divided  by  dividing  the  numer- 

(9\    ^    '    A. ^  ^    ^       *'^*^^*     Hence  we  divide  f  by  4, 

^''^9*  9x4        36        9       as  shown  in  (1),  by  dividing  the 

numerator  8  by  4. 

2.  According  to  (269 — IV),  a  fraction  is  divided  by  multiplying  the 
denominator.  Hence  we  divide  |  by  4,  as  shown  in  (2),  by  multiplying 
the  denominator  by  4,  and  reducing  the  result  to  its  lowest  terms. 

290.    Prob.  II.— To  divide  by  a  fraction. 

1.  How  many  times  is  |  of  a  given  line  contained  in  twice 
the  same  line  ? 

FIRST   STEP, 

2  lines  =  -^  of  a  line. 

5 


\  = 


SECOND  STEP. 

10  3     _     10  _ 

y  "^      5  -  ¥  -  '^=*- 


Explanation. — 1.  We  can  find  how  many  times  one  number  is  con- 
tained in  another,  only  when  both  are  of  the  same  denomination  (155). 


DIVISION     OF    FRACTIONS.  39 

Hence  we  first  reduce,  as  sliown  in  First  Step,  the  2  lines  to  10  ffthsot  a 
line ;  the  same  fractional  denomination  as  the  divisor,  3  fifths. 

2.  The  3  fifths  in  the  divisor,  as  shown  in  Second  Step,  are  contained  in 
the  10  fifths  in  the  dividend  3  times,  and  1  part  remaining,  which  makes 
^  of  a  time.     Hence  2  equal  lines  contain  |  of  one  of  them  3  J  times. 

Observe  the  following  regarding  this  solution : 
(1.)  The  dividend  is  reduced  to  the  same  fractional  denomination  as 
the  divisor  hy  multiplying  it  by  the  denominator  of  the  divisor ;  and 
when  reduced,  the  division  is  performed  by  dividing  the  numerator  of 
the  dividend  by  the  numerator  of  the  divisor. 

(2.)  By  inverting  the  terms  of  the  divisor,  these  two  operations  are 
expressed  by  the  sign  of  multiplication.  Thus,  2  -4-  ?  =  2  x  f,  which 
means  that  2  is  to  be  multiplied  by  5,  and  the  product  divided  by  3. 

2.  How  many  times  is  ^  of  a  given  line  contained  in  f  of  it  ? 

FIKST  STEP. 

2  _  4 

3  ~  6 


1  _  3 

2  ■"  6 


SECOND 

STEP. 

-T- 

3 

6 

6  6  3* 


^^^^:^^^^  ^ =  li 

Explanation. — 1.  We  reduce,  as  shown  in  First  Step,  the  dividend  f 
and  the  divisor  i  both  to  sixths  (155 — 1). 

2.  We  divide  the  f  by  J5  by  dividing  the  numerator  of  the  dividend 
by  the  numerator  of  the  divisor.  The  |  is  contained  in  f ,  as  shown  in 
Second  Step,  1\  times.     Hence  |  is  contained  1^  times  in  f. 

391.  When  dividing  by  a  fraction,  we  abbreviate  the  work 
by  inverting  the  divisor,  as  follows : 

1.  In  reducing  the  dividend  and  divisor  to  the  same  fractional  unit, 
the  product  of  the  denominators  is  taken  as  the  common  denominator, 


f^'  COMPLEX    FRACTIONS. 

and  each  numerator  is  multiplied  by  the  denominator  of  the  other  frac- 
tion, thus, 

5    ^    2  5x3    ^    2x7  15    ^    14  15      Numerator  of  dividend. 

7    *    3~7x3    *    3x7         21    *    21"'  14      Numerator  of  divisor. 

2.  By  inverting  the  divisor,  thus,  f  -5- 1  =  f  x  4  =  fl,  the  numerators 
15  and  14  are  found  at  once,  without  going  through  the  operation  of 
finding  the  common  denominator.  Hence  the  rule.  Invert  the  terms  of 
the  divisor  and  proceed  as  in  multiplication. 

294.  Pkob.  III. — To  divide  when  the  divisor  or  divi- 
dend is  a  mixed  number,  or  both. 

1.  Divide  48  by  4f . 

Explanation.— 1.   We  re- 
\^'}  4:0  -r-  4r|-  =  4:0  -7-  -^  ducc  the  divisor  4|,  as  shown 

(2.)  48-^-V-  =  48  X  ■^  =  l^      in  (1).  to  the  improper  frac- 
tion J^. 

2.  We  invert  the  divisor,  as  shown  in  (2),  according  to  (201),  and 
multiply  the  48  by  -^,  giving  lOf  as  the  quotient  of  48  divided  by  4|. 

2.  Divide  ^  by  3|. 

Explanation. — 1.  We 

(1.)  8f  -^  3 J  =  4^  -^  -^  reduce  the  dividend  and 

3  divisor,  as  shown  in  (1), 

(2.)  4^  -^  ^^1-  =  :^  X  'A:  =  V  =  2^       to    improper    fractions, 

giving  -V-  -^  V-- 
2.  We  invert  the  divisor,  y-,  as  shown  in  (2),  according  to  (291),  and 
cancel  31  in  the  numerator  63  and  denominator  31  (186),  giving  y ,'  or3f . 
Hence,  8f  -5-  3|  =  2f . 

COMPLEX    FEACTIONS. 

29T.  Certain  results  are  obtained  by  dividing  the  numera- 
tor and  denominator  of  a  fraction  by  a  number  that  is  not  an 
exact  divisor  of  each,  which  are  fractional  in  form,  but  are  not 
fractions  according  to  the  definition  of  a  fraction.  These 
fractional  forms  are  called  Complex  Fractions. 

298.  A  Complex  Fraction  is  an  expression  in  the 
form  of  a  fraction,  having  a  fraction  in  its  numerator  or 

denominator,  or  in  both  ;  thus,  |,  ^,  -|' 


COMPLEX     FRACTIONS.  41 

299.  A  Simple  Fraction  is  a  fraction  having  a  whole 
number  for  its  numerator  and  for  its  denominator. 

fbobijEms  in  complex  fractions. 

300.  Prob.  L— To  reduce  a  complex  fraction  to  a 
simple  fraction. 

Eeduce  3  to  a  simple  fraction. 

^       4f  X  12       56  Explanation.— 1.  We  find  the  least 

'if^^^  'rf%  y  12^^93  common  multiple  of  the  denominators  of 

the  partial  fractions  |  and  |,  which  is  12. 

2.  Multiplying  both  terras  of  the  complex  fraction  by  12  (235—11), 

which  is  divisible  by  the  denominators  of  the  partial  fractions,  f  and  |, 

reduces  each  term  to  a  whole  number.     4|  x  12  =  56  ;    7  J  x  12  =  93. 

41 
Therefore  ^  is  equal  to  the  simple  fraction  f |. 

303.  The  three  classes  of  complex  fractions  are  forms  of 
expressing  three  cases  of  division ;  thus, 

7.  A  mixed  number  divided  by  an  integer. 
9^.  An  integer  divided  by  a  mixed  number. 
2-|.     A  mixed  number  divided  by  a  mixed  number. 

Hence,  when  we  reduce  a  complex  fraction  to  a  simple  fraction,  as 
directed  (301),  we  in  fact  reduce  the  dividend  and  divisor  to  a  common 
denominator,  and  reject  the  denominator  by  indicating  the  division  of  the 
numerator  of  the  dividend  by  the  numerator  of  the  divisor  ;  thus, 

(3.)     5}--2f  =  ¥--f,    and    ^-M  =  «^fl  =  ll, 

the  same  result  as  obtained  by  the  method  of  multiplying  by  the  least 
C5ommon  multiple  of  the  denominators  of  the  partial  fractions. 


(!•) 

5|_ 

7  ~ 
32 

.51 

(3.) 

n" 

=  32 

(3.) 

8|_ 

=  8i 

42  REVIEW    EXAMPLES    IN   FRACTIONS. 

304.    Pkob.  II. — To  reduce  a  fraction  to  any  given 
denominator. 

1.  Exam2)les  where  the  denominator  of  the  required  fraction 
is  a  factor  of  the  denominator  of  the  given  fraction. 

Eeduce  \\  to  a  fraction  whose  denominator  is  8. 

17       17  _i_  3        5|.  Explanation. — We  observe  tliat  8,  the 

oT  -"  24  _i_'\~~  1^  denominator  of  the  required  fraction,  is  a 

factor  of  24,  the  denominator  of  the  given 

fraction.    Hence,  dividing  both  terms  of  W  by  3,  the  other  factor  of  24,  the 

5- 
fraction  is  reduced  (235 — III)  to  -r^,  a  fraction  whose  denominator  i»8. 

o 

2.  Examples  ivhere  the  denominator  of  the  required  fraction 
is  not  a  factor  of  the  denominator  of  the  given  fraction. 

Reduce  -^  to  a  fraction  whose  denominator  is  10. 

8         8    X  10        80  Explanation. — 1.  We  intro- 

(1.)     T^=T7T        in^^  T^  ^^^®  ^^^  given  denominator  10 

Id        Id  X  10        loU  ^g  ^  factor  into  the  denominator 

80    _   80   -^  13  _  6-^%      of  y\  by  multiplying,  as  shown 

^''      130  —  130  -^13  —    1(7      ^'^  ^^'  ^"*^^  terms  of  the  fraction 

by  10  (235-11). 
2.  The  denominator  130  now  contains  the  factors  13  and  10.     Hence, 
dividing  both  terms  of  the  fraction  -^^^^^  by  13  (235— III),  as  shown  in 

6  ^- 
(2),  the  result  is  — ^,  a  fraction  whose  denominator  is  10. 

REVIEW    EXAMPLES. 
307.     1.  How  many  thirtieths  in  |,  and  why?    In  f  ? 

2.  Reduce  f ,  ^^j,  j^,  /^,  and  fl  each  to  twenty-eighths. 

5    3.^ 

3.  Reduce  to  a  common  denominator  ~^,  -^,  and  /j. 

5  5x4 

4.  State  the  reason  why  .  =  — - —  (209). 

5.  Reduce  |  to  a  fraction  whose  numerator  is  12 ;  is  20 ;  is  2 ;  is  3 ; 
is  7  (235). 

6.  Reduce  to  a  common  numerator  f  and  f  (252). 

7.  Find  the  sum  of  -«,  ^■-^,  |,  f ,  and  ||. 


REVIEW   EXAMPLES    IN   FRACTIONS,  43 

8.  Find  the  value  of  (f  of  ^\  -  j\)  ^(^  +  ~^. 

9.  If  I  of  an  estate  is  worth  $3460,  what  is  |  of  it  worth  ? 
10.  $4  is  what  part  of  $8  ?    Of  $12  ?    Of  $32  ?    Of  $48  ? 
Write  the  solution  of  this  example,  with  reason  for  each  step. 

11.  If  a  man  can  do  a  piece  of  work  in  150  days,  what  part  of  it  can  he 
do  in  5  days  ?  In  15  days  ?  In  25  days  ?  In  7|  days  ?  In  3f  days  ?  In 
12|-days? 

12.  A's  farm  contains  120  acres  and  B's  380;  what  part  of  B's  farm 
isA's? 

13.  42  is  f  of  what  number  ? 

Write  the  solution  of  this  example,  with  reason  for  each  step. 

14.  $897  is  f  of  how  many  dollars  ? 

15.  f  of  76  tons  of  coal  is  j%  of  how  many  tons  ? 

16.  A  piece  of  cloth  containing  73  yards  is  |  of  another  piece.  How 
many  yards  in  the  latter? 

17.  Bought  a  horse  for  $286,  and  sold  him  for  I  of  what  he  cost ;  how 
much  did  I  lose  ? 

18.  84  is  --^^.j  of  8  times  what  number  ? 

Write  the  s:)lution  of  this  example,  with  reason  for  each  step. 

19.  A  has  $694  in  a  bank,  which  is  |  of  3  times  the  amount  B  has  in 
the  same  bank  ;  what  is  B's  money  ? 

20.  Two  men  are  86|  miles  apart ;  when  they  meet,  one  has  traveled 
8 J  miles  more  than  the  other  ;  how  far  has  each  traveled  ? 

21.  If  j\  of  a  farm  is  valued  at  $4732 1,  what  is  the  value  of  the 
whole  farm  ? 

22.  The  less  of  two  numbers  is  432f,  and  their  difference  123 j^'^.  Find 
the  greater  number. 

23.  A  man  owning  |  of  a  saw-mill,  sold  f  of  his  share  for  $2800 ; 
what  was  the  value  of  the  mill  ? 

24.  What  number  diminished  by  f  and  f  of  itself  leaves  a  remainder 
of  32? 

25.  I  put  *  of  my  money  in  the  bank  and  gave  f  of  what  I  had  left  to 
a  friend,  and  had  still  remaining  $400.     How  much  had  I  at  first  ? 

26.  Sold  342  bushels  of  wheat  at  $1|  a  bushel,  and  expended  the 
amount  received  in  buying  wood  at  $4|  a  cord.  How  many  cords  of 
wood  did  I  purchase  ? 

27.  If  f  of  4  pounds  of  tea  cost  $2 J,  how  many  pounds  of  tea  can  be 
bought  for  $7 i  ?    For  $12.}?    For^^^^? 


44         REVIEW     QUESTIONS     ON    FRACTIONS, 

28.  If  5  be  added  to  both  terms  of  the  fraction  2,  how  much  will  its 
value  be  changed,  and  why  ? 

29.  I  exchanged  47f  bushels  of  corn,  at  $|  per  bushel,  for  24|  bushels 
of  wheat ;  how  much  did  the  wheat  cost  a  bushel  ? 

30.  A  can  do  a  piece  of  work  in  5  days,  B  can  do  the  same  work  in 
7  days  ;  in  what  time  can  both  together  do  it  ? 

31.  Bought  I  of  844  acres  of  land  for  |  of  $3584^- ;  what  was  the  price 
per  acre? 

32.  A  boy  while  fishing  lost  |  of  his  line  ;  he  then  added  8  feet,  which 
was  I  of  what  he  lost;  what  was  the  length  of  the  line  at  first?      ^^ 

33.  A  merchant  bought  a  quantity  of  cloth  for  $2849 1,  and  sold  it  for 
y'^  of  what  it  cost  him,  thereby  losing  $?  a  yard.  How  many  yards  did 
he  purchase,  and  at  what  price  per  yard  ? 

34.  A  tailor  having  276|  yards  of  cloth,  sold  f  of  it  at  one  time  and  f 
at  another ;  what  is  the  value  of  the  remainder  at  $3  a  yard  ? 

35.  A  man  sold  /^  of  his  farm  at  one  time,  {  at  another,  and  the 
remainder  for  $180  at  $45  an  acre  ;  how  many  acres  were  there  in 
the  farm  ? 

36.  A  merchant  owning  ||  of  a  ship,  sells  ^  of  his  share  to  B,  and  f 
of  the  remainder  to  C  for  $600|^ ;  what  is  the  value  of  the  ship  ? 


KEVIEW  AND  TEST   QUESTIONS. 

308.  1.  Define  Fractional  Unit,  Numerator,  Denominator,  Im- 
proper Fraction,  Reduction,  Lowest  Terms,  Simple  Fraction,  Common 
Denominator,  and  Complex  Fraction. 

2.  What  is  meant  by  the  unit  of  a  fraction  ?  Illustrate  by  an  example. 

3.  When  may  |  be  greater  than  ^  ?    \  than  \  ? 

4.  State  the  three  principles  of  Reduction  of  Fractions,  and  illustrate 
each  by  lines. 

5.  Illustrate  with  lines  or  objects  each  of  the  following  propositions  : 

I.  To  diminish  the  numerator,  the  denominator  remaining  the 

same,  diminishes  the  value  of  the  fraction. 
II.  To  increase  the  denominator,  the  numerator  remaining  the 
same,  diminishes  the  value  of  the  fraction. 

III.  To  increase  the  numerator,  the  denominator  remaining  the 

same,  increases  the  value  of  the  fraction. 

IV.  To  diminish  the  denominator,  the  numerator  remaining  the 

same,  increases  the  value  of  the  fraction. 


^%^  '        :---=7 


DECIMAL     FRACTIONS.  45 

6.  What  is  meant  by  tlie  Least  Common  Denominator  ? 

7.  When  the  denominators  of  the  given  fractions  are  prime  to  each 
other,  how  is  the  Least  Common  Denominator  found,  and  why  ? 

8.  State  the  five  problems  in  reduction  of  fractions,  and  illustrate  each 
by  the  use  of  lines  or  objects. 

9.  Show  that  multiply- ing  the  denominator  of  a  fraction  by  any  num- 
ber divides  the  fraction  by  that  number  (269). 

10.  Show  by  the  use  of  lines  or  objects  the  truth  of  the  following  : 

(1.)  i  of  2  equals  f  of  1.  (3.)  ^  of  5  equals  |  of  1. 

(3.)  f  of  1  equals  ^  of  3.  (4.)  -f  of  9  equals  4  times  \  of  9. 

11.  To  give  to  another  person  f  of  14  silver  dollars,  how  many  of  the 
dollar-pieces  must  you  change,  and  what  is  the  largest  denomination  of 
change  you  can  use  ? 

12.  Show  by  the  use  of  objects  that  the  quotient  of  1  divided  by  a 
fraction  is  tlie  given  f motion  inverted. 

13.  Why  is  it  impossible  to  perform  the  operation  in  |  +  |,  or  in  f— f, 
without  reducing  the  fractions  to  a  common  denominator  ? 

14.  Why  do  we  invert  the  divisor  when  dividing  by  a  fraction? 
Illustrate  your  answer  by  an  example. 

15.  What  objection  to  calling  —  a  fraction  (226)  ? 

16.  State,  and  illustrate  with  lines  or  objects,  each  of  the  three  classes 
of  so-called  Complex  Fractions. 

17.  Which  is  the  greater  fraction,  ^  or  f  J,  and  how  much? 

18.  To  compare  the  value  of  two  or  more  fractions,  what  must  be 
done  with  them,  and  why  ? 

Qi  4     2i  3      5^  2- 

19.  Compare  ~  and  -  ;  -^  and  —  ;  -|  and  -j,  and  show  in  each  case 

7  8      9  10     7i  o^ 

which  is  the  greater  fraction,  and  how  much. 

20.  State  the  rule  for  working  each  of  the  following  examples : 

(1.)  3|  +  4f  +  8|. 

(2.)  (7f  +  5^)  -  (8  -  21). 

(3.)  5  X  I  of  I  of  27. 

(4.)  8|  X  5|. 

(5.)  I  X  f .    Explain  by  objects. 

(6.)  YS-i-^,    Explain  by  objects. 

21.  Illustrate  by  an  example  the  application  of  Cancellation  in  multi- 
plication and  division  of  fractions. 


4f6  DECIMAL     FRACTIONS. 

DECIMAL    FEAOTIOE"S. 
DEFINITIONS. 

309.   A  unit  is  separated  into  decimal  parts  when  it 

is  divided  into  teiiths  ;  thus. 


DECIMAL  PARTS. 


310.  A  Decimal  Fractional  Unit  is  one  of  the 

decimal  parts  of  anything. 

311.  By  making  a  whole  or  ^mit  into  decimal  parts,  and 
one  of  these  parts  into  decimal  parts,  and  so  on,  we  obtain  a 
series  of  distinct  orders  oi  decimal  fractional  tmits,  each  -^^  of 
the  preceding,  having  as  denominators,  respectively,  10,  100, 
1000,  and  so  on. 

Thus,  separating  a  whole  into  decimal  parts,  we  have,  according  to 
(246),  1  =  {%;  making  J^  into  decimal  parts,  we  have,  according  to 
(252),  ^V  =  TTH7 ;  in  tlie  same  manner,  ^-^^  =  ^-^,  ^-^^o  =  WriViy  and 
so  on.  Hence,  in  the  series  of  fractional  units,  y^^^,  j^^,  j  \^^,  and  so  on, 
each  unit  is  one-tenth  of  the  preceding  unit. 

312.  A  Decimal  Fraction  is  a  fraction  whose  denom- 
inator is  10,  100,  1000,  etc.,  or  1  with  any  number  of  ciphers 
annexed.     Thus,  -^,  yj^,  yjfoj  ^^'^  decimal  fractions. 

31.3.  The  Decimal  Sir/n  (.),  called  the  decirnal  pointy 
is  used  to  express  a  decimal  fraction  without  writing  the 
denominator,  and  to  distinguish  it  from  an  integer. 

315,  A  flecimaJ  fraction  expressed  without  writing 
the  denominator  h  called  simply  a  Decimal. 

Tlius,  wc  speak  of  .79  as  the  (Primal  seventy-nine,  yet  we  mean  the 
decimal  fraction  seventy-nine  hundredths. 


LECIMAL     FRACTIONS,  47 


NOTATION    AND    NUMERATION. 

314.  Prop.  I. — A  decimal  fraction  is  expressed  without 
writing  the  denominator  by  using  the  decimal  point,  and 
placing  the  numerator  at  the  right  of  the  period. 

Thus,  -^-Q  is  expressed  .7 ;  {^^^  is  expressed  .35. 

316.  Prop.  II. — Ciphers  at  the  left  of  significant  figures 
do  not  iiicrease  or  diminish  the  member  expressed  by  these 
figures. 

Thus,  0034  is  thirty-four,  the  same  as  if  written  34  without  the  two 
ciphers. 

317.  Prop.  111.— When  the  fraction  in  a  mixed  number 
is  expressed  decimally,  it  is  written  after  the  integer,  with 
the  decimal  point  between  them. 

Thus,  57  and  .09  are  written  57.09  ;  8  and  .0034  are  written  8.0034. 

320.  Prop.  IV. — Every  figure  in  the  numerator  of  a  deci- 
mal fraction  represents  a  distinct  order  of  decimal  units. 

Thus,  tVoV  is  equal  to  ^Wif  +  t^^  +  nrW-  But,  according  to  (255), 
■f^^j^  =  ■^^,  and  yf ^^  =  yf  ^.  Hence,  5,  3,  and  7  each  represent  a  distinct 
order  of  decimal  fractional  units,  and  {'^^^,  or  .537  may  be  read  5  tenths 
3  hundredths  and  7  thousandtlis. 

321.  Prop.  V. — A  decimal  is  read  correctly  by  reading  it 
as  if  it  were  an  integer  and  giving  the  name  of  the  right-hand 
order. 

Thus,  .975  =  ^"o"^  +  y^^  +  -n/VTr*  Hence  is  read,  nine  hundred 
seventy-five  thousandths. 

1 .  Observe  that  when  there  are  ciphers  at  the  left  of  the  decimal, 
according  to  (316),  they  are  not  regarded  in  reading  the  number ;  thus, 
.062  is  read  sixty-two  thousandths. 

2.  The  name  of  the  lowest  order  is  found,  according  to  (314),  by 
prefixing  1  to  as  many  ciphers  as  there  are  figures  in  the  decimal.  For 
example,  in  .00209  there  are  five  figures  ;  hence  the  denominator  is  1 
with  five  ciphers  ;  thus,  100000,  read  handred-thousandths. 


48  DECIMAL     FRACTIONS: 

REDUCTION. 

PBEPARATOltT     l*JiOrOSITION8, 

The  following  preparatory  propositions  should  be  ver^ 
carefully  studied. 

325.  Prop.  I. — Annexing  a  cipher  or  multiplying  a  num- 
ber hy  10  introduces  into  the  number  the  tivo  prime  factors  2 
and  5, 

Thus,  10  being  equal  2  x  5,  7  x  10  or  70  =  7  x  (2  x  5).  Hence  a  number 
must  contain  2  and  5  as  a  factor  at  least  as  many  times  as  there  are 
ciphers  annexed. 

326.  Prop.  II.— ^  fraction  in  its  lowest  terms,  whose 
denomi7iator  contains  no  other  prime  factors  than  2  or  5, 
can  be  reduced  to  a  simple  decimal. 

Observe  that  every  cipher  annexed  to  the  numerator  and  denominator 
makes  each  divisible  once  by  2  and  5  (325).  Hence,  if  the  denominator 
of  the  given  fraction  contains  no  other  factors  except  2  and  5,  by  annex- 
ing ciphers  the  numerator  can  be  made  divisible  by  the  denominator, 
and  the  fraction  reduced  to  a  decimal. 

Thus,  I  =  i^^-o  (235—11).  Dividing  both  terms  of  the  fraction  by  8 
(235-III),  we  have  f^^  =  ^^^  =  .875. 

327.  Prop.  III. — A  fraction  in  its  lowest  terms,  whose 
denominator  contains  any  other  prime  factors  than  2  or  5,  can 
be  reduced  only  to  a  complex  decimal. 

Observe  that  in  this  case  annexing  ciphers  to  the  numerator  and 
denominator,  which  (325)  introduces  only  the  factors  2  and  5,  cannot 
make  the  numerator  divisible  by  the  given  denominator,  which  contains 
other  prime  factors  than  2  or  5. 

Hence,  a  fraction  will  remain  in  the  numerator,  after  di^iding  the 
numerator  and  denominator  by  the  denominator  of  the  given  fraction, 
however  far  the  division  may  be  carried. 

Thus,  ^\  =  ^l^n  (235—11).    Dividing  both  numerator  and  denom- 

inator  by  31,  we  have  =  Inoo  ~  -^^^H*  ^  complex  decimal. 


DECIMAL     FRACTIONS.  49 

328.  Pkop.  IV. — The  same  set  of  figures  must  recur  in-' 
definitely  in  the  same  order  in  a  complex  decimal  which  ca?mot 
he  reduced  to  a  simple  decimal. 

_         7        700Q0        6363yV 

T^^^'  li  =  IIoooo  =  loooo"  =  -^^^^A- 

Observe  carefully  the  following: 

1.  In  any  division,  the  number  of  different  remainders  that  can  occur 
is  1  less  than  the  number  of  units  in  the  divisor. 

Thus,  if  5  is  the  divisor,  4  must  be  the  greatest  remainder  we  can 
have,  and  4,  3,  3,  and  1  are  the  only  possible  different  remainders  ;  hence, 
if  the  division  is  continued,  any  one  of  these  remainders  may  recur. 

2.  Since  in  dividing  the  numerator  by  the  denominator  of  the  given 
fraction,  each  partial  dividend  is  formed  by  annexing  a  cipher  to  the 
remainder  of  the  previous  division,  when  a  remainder  recurs  the  partial 
dividend  must  again  be  the  same  as  was  used  when  this  remainder 
occurred  before ;  hence  the  same  remainders  and  quotient  figures  must 
recur  in  the  same  order  as  at  first. 

3.  If  we  stop  the  division  at  any  point  where  the  given  numerator 
recurs  as  a  remainder,  we  have  the  same  fraction  remaining  in  the  nu- 
merator of  the  decimal  as  the  fraction  from  which  the  decimal  is  derived. 

Thus  7_700_63t'._ 

7        70000       6363W        noao  ■,        a 

"'      u  =  noooo  ==  -mm  ==  •  ''^'^^'^'  ''"'^  ^^  °°- 

329.  Prop.  V. — The  value  of  a  fraction  luhich  can  only 
he  reduced  to  a  complex  decimal  is  expressed,  nearly,  as  a  sim- 
ple decimal,  hy  rejecting  the  fraction  from  tlie  numerator. 

3       27  ^ 
Thus,  r-j-  =  -j^  (327).    Rejecting  the  ^  from  the  numerator,  we 

have  y^^,  a  simple  fraction,  which  is  only  y\  of  -^  smaller  than  the 

27tt 
given  fraction  y\  or  -~^  • 
100 

Observe  the  following : 

2.  By  taking  a  sufficient  number  of  places  in  the  decimal,  the  true 
value  of  a  complex  decimal  can  be  expressed  so  nearly  that  what  is  re- 
jected is  of  no  consequence. 

3 


so  DECIMAL     FRACTIONS. 

^,         3       27272727A       .      . 

'  11  ~  TOOOOOOOO  '  ^^J®^*"^&  *^®  h  from  tlie  numerator,  we  have 

■NmoWVuj  or  .27272727,  a  simple  decimal,  which  is  only  -^j  of  1  hundred- 
millionths  smaller  than  the  given  fraction. 

2.  The  approximate  value  of  a  complex  decimal  which  is  expressed  by 
rejecting  the  given  fraction  from  its  numerator  is  called  a  Circulating 
Becimaly  because  the  same  figure  or  set  of  figures  constantly  recur. 

330.  Prop.  VI. — Diminishing  ilie  numerator  and  denom- 
inator hj  the  same  fractional  part  of  each  does  not  change  ihe 
value  of  a  fraction. 

Be  particular  to  master  the  following,  as  the  reduction  of  circulating 
decimals  to  common  fractions  depends  upon  this  proposition. 

1.  The  truth  of  the  proposition  may  be  shown  thus : 

^_9^-|of    9_^-3_6_3 

12  ~  12  -  i  of  12  ~  12  -  4  ~  8  ~  4 
Observe  that  to  diminish  the  numerator  and  denominator  each  by  i  of 
itself  is  the  same  as  multiplying  each  by  f .  But  to  multiply  each  by  I , 
we  multiply  each  by  2  (235 — II),  and  then  divide  each  by  3  (235— III), 
which  does  not  change  the  value  of  the  fraction  ;  hence  the  truth  of  the 
proposition. 

2.  From  this  proposition  it  follows  that  the  value  of  a  fraction  is  not 
changed  by  subtracting  1  from  the  denominator  and  the  fraction  itself 
from  the  numerator. 

3      3 3      2?^ 

Thus,  ^  =  =      ^  =-^'     Observe  that  1  is  the  I  of  the  denominator  5, 
0       5  —  1        4 

and  I  is  \  of  the  numerator  3 ;  hence,  the  numerator  and  denominator 

being  each  diminished  by  the  same  fractional  part,  the  value  of  the 

fraction  is  not  changed. 

DEFINITIONS. 

331.  A  Simple  Decimal  is  a  decimal  wbose  numerator 
is  a  whole  number  ;  thus,  -f^  or  .93. 

Simple  decimals  are  also  called  Mnite  Decimals. 

332.  A  Complex  Decimal  is  a  decimal  whose  numer- 
ator is  a  mixed  number;  as  ~^^  or  .26|. 


DECIMAL     FRACTIONS,  51 

There  are  two  classes  of  complex  decimals : 

1.  Those  whose  value  can  be  expressed  as  a  simple  decimal  (326),  as 
.23^  =  .235  ;  .32|  =  .3275. 

2.  Those  whose  value  cannot  be  expressed  as  a  simple  decimal  (327), 
as  .53^  —  .53333  and  so  on,  leaving,  however  far  we  may  carry  the  deci- 
mal places,  I  of  1  of  the  lowest  order  unexpressed.     See  (328j. 

333.  A  Circalating  Decimal  is  an  approximate 
value  for  a  complex  decimal  which  cannot  he  reduced  to  a  sim- 
ple decimal. 

Thus,  .0G6  is  an  approximate  value  for  .666|  (329). 

334.  A  Hepetend  is  the  figure  or  set  of  figures  that  are 
repeated  in  a  circulating  decimal. 

335.  A    Clrculafinf/  Decimal  is  exiwessed  by 

writing  the  repetend  once.  When  the  repetend  consists  of 
one  figure,  a  point  is  placed  over  it;  when  of  more  than  one 
figure,  points  are  placed  over  the  first  and  last  figures;  thus, 
.333  and  so  on,  and  .592592+  are  written  .3  and  .592. 

336.  A  Pare  Cifculatinff  Decimal  is  one  which 
commences  with  a  repetend,  as  .8  or  .394. 

337.  A   Mixed  Circulatincf   Decimal  is  one  in 

which  the  repetend  is  preceded  by  one  or  more  decimal  places, 
called  the  finite  part  of  the  decimal,  as  .73  or  .004725,  in 
which  .7  or  .004  is  the  finite  part. 

jltjTjsthation    of   process. 

338.  Prob.  I. — To  reduce  a  commoii  fraction  to  a 
decimal. 

Reduce  |  to  a  decimal. 

3  _  3000  _   375  Explanation.  —  1.    We   annex 

8        8000        1000  ^^  the  same  number  of  ciphers  to  both 

terms  of  the  fraction  (235—11), 
and  divide  the  resulting  terms  by  8,  the  significant  figure  in  the  denom- 
inator which  must  give  a  decimal  denominator.  Hence,  |  expressed 
decimally  is  .375. 


6'Z  DECI3IAL     FRACTIONS, 

2.  In  case  annexing  ciphers  does  not  make  the  numerator  divisible 
(327)  by  the  significant  figures  in  the  denominator,  the  number  of  places 
in  the  decimal  can  be  extended  indefinitely. 

In  practice,  we  abbreviate  the  work  by  annexing  the  ciphers  to  the 
numerator  only,  and  dividing  by  the  denominator  of  the  given  fraction, 
pointing  off  as  many  decimal  places  in  the  result  as  there  were  ciphers 
annexed. 

341.  Pkob.  II. — To  reduce  a  simple  decimal  to  a 
eoiiiinou  fraction. 

Eeduce  .35  to  a  common  fraction. 

35  7  Explanation. — We  write  the  decimal  with 

•^^  ^^  Tqq  ^^  20  ^^'^    denominator,    and    reduce    the    fraction 

(255)  to  its  lowest  terms 

344,  Prob.  III. — To  find  tlie  true  value  of  a  pure 
circulating-  decimal. 

Find  the  true  value  of  .72. 

..72  72  72  8  EXPLANATION.-In  tak- 

.72  = = =  —  =  —         ijiff  -72  as  the  approximate 

1 nn       1  no       i        qq       i i 

x\j\j        xuu        X         xjv        j.±  mlue  of  a  given  fraction, 

we  have  subtracted  the  given  fraction  from  its  own  numerator,  as  shown 

in  (329 — V).     Hence,  to  find  the  true  value  of  y^^,  we  must,  according 

to  (330 — VI,  2),  subtract  1  from  the  denominator  100,  which  makes  the 

denominator  as  many  9's  as  there  are  places  in  the  repetend. 

347.  Prob.  IV. — To  find  the  trvie  value  of  a  mixed 
circulating  decimal. 

Find  the  true  value  of  .318. 

(1.)     .3i8  =  .3H  =  f  =  S.|. 

Explanation. — 1.  We  find,  according  to  (344),  the  true  value  of 

the  repetend  .018,  which  is  .0J|.     Annexing  this  to  the  ,3,  the  finite  part, 

we  have  .3^§^.  the  true  value  of  .318  in  the  form  of  a  complex  decimal. 

gi  ft 
2.  We  reduce  the  complex  decimal  .3.^^,  or  ~^^  to  a  simple  fraction 

by  multiplying,  according  to  (300),  both  terms  of  the  fraction  by  99, 

318 
giving  -5^  =  m  =  /jj.    Hence  the  true  value  of  .318  is  ^V 


DECIMAL     FRACTIONS,  53 

Abbreviated  Solution.— Observe 
(2.)      .318      Given  decimal.  gi  s 

_3      Finite  part.  *^^*  ^^  simplifying  -j^,  we  multiplied 

315      315  — -    7_,  T^oth  terms  by  99.     Instead  of  raulti- 

^  ^^        ^^*  plying  tlie  3  by  99,  we  may  multiply 

by  100  and  subtract  3  from  the  product.     Hence  we  add  tlie  18  to  300, 

and  subtract  3  from  the  result,  which  gives  us  the  true  numerator. 

To  abbreviate  the  work  ; 

From  the  given  decimal  subtract  the  finite  part  for  a  numer- 
ator, and  for  a  denominator  write  as  majiy  9^s  as  there  are 
figures  in  the  repetend,  with  as  maiiy  ciphers  annexed  as  there 
are  figures  in  the  finite  part. 


ADDITION. 

IliljVSTJtATION     OF     PROCESS. 

351.  Find  the  sum  of  34.8,  6.037,  and  27.62. 

Explanation. — 1.  We  arrange 
the  numbers  so  that  units  of  the 
same  order  stand  in  the  same 
column. 

2.  We  reduce  the  decimals  to  a 
common  denominator,  as  shown  in 
(1),  by  annexing  ciphers. 
3.  We  add  as  in  integers,  placing  the  decimal  point  before  the  tenths 
in  the  sum. 

SUBTRACTION. 
353.  Find  the  difference  between  83.7  and  45.392. 

g  3    7  0  0  Explanation. — 1.  We  arrange  the  numbers  so 

/L  ^    q  Q  9  *hat  units  of  the  same  order  stand  in  the  same  column. 

'- — '- —  2.    We  reduce   the  decimals,   or  regard  them  as 

3  8.308  reduced  to  a  common  denominator,  and  then  subtract 

as  in  whole  numbers. 
The  reason  of  this  course  is  the  same  as  given  in  addition.     The 
ciphers  are  also  usually  omitted. 


(1-) 

(2.) 

34.800 

34.8 

6.037 

6.037 

27.620 

27.62 

68.457 

68.457 

54  DECIMAL     FRACTIONS. 

MULTIPLICATION. 
355.  Multiply  3.27  by  8.3. 

(1.)  3.27  X  8.3  =  ^  x^ 

327      83_a7141_ 

^^■>      loo""  10-  1000  -^'-^^i 

Explanation. — 1.  Observe  that  3.27  and  8.3  are  mixed  numbers; 
hence,  according  to  (282),  they  are  reduced  before  being  multiplied  to 
improper  fractions,  as  shown  in  (1). 

2.  According  to  (270),  ff^  x  ff,  as  shown  in  (2),  equals  27.141. 
Hence  27.141  is  the  product  of  3.27  and  8.3. 

The  work  is  abbreviated  thus : 

(3.) 

q    n  w  We  observe,  as  shown  in  (2),  that  the  product  of 

*  3.27  and  8.3  must  contain  as  many  decimal  places  as 

. ^  •  ^  there  are  decimal  places  in  both  numbers.     Hence 

9  81  ^ve  multiply  the  numbers  as  if  integers,  as  shown  in 

2  (5  2  6  ^^'  ^^^  point  off  in  the  product  as  many  decimal 

—  places  as  there  are  decimal  places  in  both  numbers. 

27.141 

DIVISION. 
pmeparatout   propositions, 

358.  Prop.  I. —  Wlien  the  divisor  is  greater  than  the  divi- 
dend, the  quotient  expresses  the  part  the  dividend  is  of  the 
divisor. 

Thus,  4  H-  6  =  f  =  |.  The  quotient  f  expresses  the  part  the  4 
is  of  6. 

1.  Observe  that  the  process  in  examples  of  this  kind  consists  in 
reducing  the  fraction  formed  by  placing  the  divisor  over  the  dividend 
to  its  lowest  term^.  Thus,  32 -?- 56  =  f  f ,  which  reduced  to  its  lowest 
terms  gives  :^. 

2-  In  case  the  result  is  to  be  expressed  decimally,  the  process  then 
consists  in  reducing  to  a  decimal  according  to  (338),  the  fraction 
formed  by  placing  the  dividend  over  the  divisor.  Thus,  5  -5-  8  =  |, 
reduced  to  a  decimal  equals  .625. 


DECIMAL     FRACTIONS,  55 

359.  Prop.  II. — The  fraction  remaining  after  the  division 
of  one  i7iteger  by  another  expresses  the  part  the  remainder  is 
of  the  divisor. 

Tims,  42  -r- 11  =  3i\.  The  divisor  11  is  contained  3  times  in  42  and 
9  left,  wliicli  is  9  parts  or  y\  of  Ihe  divisor  11.  Hence  we  say  that  the 
divisor  11  is  contained  3^^^  times  in  42.  We  express  the  -^j  decimally 
by  reducing  it  according  to  (338).     Hence,  S^^  =  3.81. 

360.  Prop.  III. — Divisio7i  is  possible  only  when  the  divi- 
defid  and  divisor  are  both  of  the  same  denomination  (155 — I). 

For  example,  f^  -^  y§^,  or  .3  -f-  .07  is  impossible  until  the  dividend 
and  divisor  are  reduced  to  the  same  fractional  denomination  ;  thus, 
.3  -J-  .07  =  .30  -V-  .07  =  4|  =  4.285714. 

ILLUSTRATION     OF     P M O C  E S  S  , 

361.  Ex.  1.  Divide  .6  by  .64. 

(1.)  .8^.64  =  .60-^■.  64 

(2.)  60-^64  =  —  =  .9375 

Explanation. — 1.  We  reduce,  as  shown  in  (1),  the  dividend  and 
divisor  to  the  same  decimal  unit  or  denomination  (200). 

2.  We  divide,  according  to  (200),  as  shown  in  (2),  the  numerator  60 
by  the  numerator  64,  which  gives  f  J.  Reducing  ff  to  a  decimal  (338), 
we  have  .6  -4-  .64  ==  .9375. 

Ex.  2.  Divide  .63  by  .0022. 

(1.)         .63-^. 0022  =  . 6300-T-. 0022 
(2.)        6300-T-22  =  286^  =  386.36 

Explanation. — 1.  We  reduce,  as  shown  in  (1),  the  dividend  and 
divisor  to  the  same  decimal  unit  by  annexing  ciphers  to  the  dividend 
(850). 

2.  We  divide,  according  to  (290),  as  shown  in  (2),  the  numerator 
6300  by  the  numerator  22,  givinpr  as  a  quotient  286y\. 

3.  We  reduce,  according  to  (338),  the  y\  in  the  quotient  to  a  decimal, 
giving  the  repetend  .36.     Hence,  .63  -*-  .0022  =  286.36. 


i 

DECIMAL 

FRACTIONS 

Ex.3. 

Divide  16.821 

by  2.7. 

(1.) 

16. 

821  -V- 

2.7  = 

16 

.821-T-2 

(2.) 

16. 

821  -^ 

2.700  = 

16821 
1000    * 

(3.)     2  7 1  0  0  )  1  6  8| 

21(6. 

23 

EXPLAI 

162 

duce,  as 

81 


56 


J. 700 
2700 
'   1000 

isfATiON. — 1.  We  re- 
shown    in   (1),  the 

dividend  and  divisor  to  the 

"  '^  same  decimal  unit  by  annex. 

5  4  ing    ciphers    to    the    divisor 

"~g"Y  (350). 

2.  The  dividend  and  divisor 
each  express  thousandths  as 
shown     in    (2).       Hence    we 
reject  the  denominators  and  divide  as  in  integers  (1290). 

3.  Since  there  are  ciphers  at  the  right  of  the  divisor,  they  may  be  cut 
off  by  cutting  off  the  same  number  of  figures  at  the  right  of  the  dividend 
(142).  Dividing  by  37,  we  find  that  it  is  contained  6  times  in  1G8,  with 
G  remaining. 

4.  The  6  remaining,  with  the  two  figures  cut  off,  make  a  remainder  of 
621  or  /7V5.  This  is  reduced  to  a  decimal  by  dividing  both  terms  by  27. 
Hence,  as  shown  in  (3),  we  continue  dividing  by  27  by  taking  down  the 
two  figures  cut  off. 

The  work  is  ubbreyiated  thus: 

We  reduce  the  dividend  and  divisor  to  the  same  decimal  unit  by  cut- 
ting off  from  the  right  of  the  dividend  the  figures  that  express  lower 
decimal  units  than  the  divisor.  We  then  divide  as  shown  in  (3),  prefix- 
ing the  remainder  to  the  figures  cut  off  and  reducing  the  result  to  a 
decimal. 

REVIEW    EXAMPLES. 

364.  Answers  involving  decimals,  unless  otherwise  stated, 
are  carried  to  four  decimal  places. 

Reduce  each  of  the  following  examples  to  decimals : 
8.    U. 


11 

31 

14. 

(3H-i)xl 

7^ 

8 

12. 

foflf. 

15. 

f  of  .3 

13. 

5A-5|. 

8|  -  4.3 

10.    ^. 

20.  Four  loads  of  hay  weighed  respectively  2583.07,  3007? ,  2567|,  and 
3074}^^  pounds ;  what  was  the  total  weight  ? 


DECIMAL     FRACTIO  NS.  57 

22.  At  $1.75  per  100,  what  is  the  cost  of  5384  oranges  ? 

24.  If  freight  from  St.  Louis  to  New  York  is  $.39*  per  100  pounds, 
what  is  the  cost  of  transporting  3  boxes  of  goods,  weighing  respectively 
7831,  32o|,  and  288|  pounds? 

25.  A  piece  of  broadcloth  cost  $195. 38 1,  at  f3.27  per  yard.  How 
many  yards  does  it  contain  ? 

26.  A  person  having  $1142.49|,  wishes  to  buy  an  equal  number  of 
bushels  of  wheat,  corn,  and  oats ;  the  wheat  at  |1.37,  the  corn  at  $.87|, 
and  the  oats  at  $.35|.     How  many  bushels  of  each  can  he  buy  ? 

20.  A  produce  dealer  exchanged  48|  bushels  oats  at  39|  cts.  per 
bushel,  and  13 1  barrels  of  apples  at  $3.85  a  barrel,  for  butter  at  37|  cts. 
a  pound  ;  how  many  pounds  of  butter  did  he  receive  ? 

30.  A  grain  merchant  bought  1830  bushels  of  wheat  at  $1.25  a 
bushel,  570  bushels  corn  at  784^  cts.  a  bushel,  and  468  bushels  oats  at 
35f  cts.  a  bushel.  He  sold  the  wheat  at  an  advance  of  11  \  cts.  a  bushel, 
the  corn  at  an  advance  of  9|  cts.  a  bushel,  and  the  oats  at  a  loss  of  3  cts. 
a  bushel.  How  much  did  he  pay  for  the  entire  quantity,  and  what  was 
his  gain  on  the  transaction  ? 

EEVIEW    AND    TEST    QUESTIONS. 

365.  1.  Define  Decimal  Unit,  Decimal  Fraction,  Repetend,  Cir- 
culating Decimal,  Mixed  Circulating  Decimal,  Finite  Decimal,  and 
Complex  Decimal. 

2.  In  how  many  ways  may  |  be  expressed  as  a  decimal  fraction, 
and  why? 

3.  What  effect  have  ciphers  written  at  the  left  of  an  integer  ?  At  the 
left  of  a  decimal,  and  why  in  each  case  (316)  ? 

4.  Show  that  each  figure  in  the  numerator  of  a  decimal  represents  a 
distinct  order  of  decimal  units  (320). 

5.  How  are  integral  orders  and  decimal  orders  each  related  to  the  units 
(323)  ?    Illustrate  your  answer  by  lines  or  objects. 

6.  Why  in  reading  a  decimal  is  the  lowest  order  the  only  one  named  ? 
Illustrate  by  examples  (321). 

7.  Give  reasons  for  not  regarding  the  ciphers  at  the  left  in  reading 
the  numerator  of  the  decimal  ,000403. 

8.  Reduce  l  to  a  decimal,  and  give  a  reason  for  each  step  in  the  process. 

9.  When  expressed  decimally,  how  many  places  must  J^\  give,  and 
why  ?    How  many  must  ^'^  give,  and  why  ? 

10.  Illustrate  by  an  example  the  reason  why  ^\  cannot  be  expressed 
as  a  simple  decimal  (327). 


58  DECIMAL     FRACTIONS. 

11.  State  wiiat  fractions  can  and  what  fractions  cannot  be  expressed 
as  simjile  decimals  (31i5(>  and  3ii7).     Illustrate  by  examples. 

12.  In  reducing  f  to  a  complex  decimal,  why  must  the  numerator  5 
recur  as  a  remainder  (r528  — 1  and  2j  ? 

13.  Show  that,  according  to  (ii35 — II  and  III),  the  value  of  ^|  will 
not  be  changed  if  we  diminish  the  numerator  and  denominator  each  by 
I  of  itself. 

14.  Show  that  multiplying  9  by  I5  increases  the  9  by  |  of  itself. 

15.  Multiplying  the  numerat(n-  and  denominator  of  ^|  each  by  1| 
produces  what  change  in  the  fraction,  and  why  ? 

16.  Show  that  in  diminishing  the  numerator  of  |  by  f  and  the 
denominator  by  1  we  diminish  each  by  the  same  part  of  itself. 

17.  In  taking  .3  as  the  value  of  3,,  what  fraction  has  been  rejected 
from  the  numerator?  What  must  be  rejected  from  the  denominator  to 
make  .3  =  -J,  and  why  V 

18.  Show  that  the  true  value  of  .81  is  f  J^.  Give  a  reason  for  each 
step. 

19.  Explain  the  process  of  reducing  a  mixed  circulating  decimal  to 
a  fraction.    Give  a  reason  for  each  step. 

20.  How  much  is  .33333  less  than  i  and  why? 

21.  How  much  is  .571428  less  than  |,  and  why? 

22.  Find  the  sum  of  .73,  .0049,  .089,  6.58,  and  9.08703,  and  explain 
each  step  in  the  process  (201 — I  and  II). 

23.  If  tenths  are  multiplied  by  hundredths,  how  many  decimal  places 
will  there  be  in  the  product,  and  why  (3155)  ? 

24.  Show  that  a  number  is  multiplied  by  10  by  moving  the  decimal 
point  one  place  to  the  right ;  by  100  by  moving  it  two  places  ;  by  1000 
three  places,  and  so  on. 

25.  State  a  rule  for  pointing  off  the  decimal  places  in  the  product  of 
two  decimals.     Illustrate  by  an  example,  and  give  reasons  for  your  rule. 

26.  Multiply  385.28  by  .742,  multiplying  first  by  the  4  hundredths, 
then  by  the  7  tenths,  and  last  by  the  2  tJwusandths. 

27.  Why  is  the  quotient  of  an  integer  divided  by  a  proper  fraction 
greater  than  the  dividend  ? 

28.  Show  that  a  number  is  divided  by  10  by  moving  the  decimal  point 
one  place  to  the  left ;  by  100  by  moving  it  two  places  ;  by  1000,  three 
places  ;  by  10000,  four  places,  and  so  on. 

29.  Divide  4.9  by  1.305,  and  give  a  reason  for  each  step  in  the  process. 
Carry  the  decimal  to  three  places. 

30.  Give  a  rule  for  division  of  decimals. 


DEX0  3IINATE     NUMBERS.  59 

DElsrOMIKATE     l^UMBEES. 
DEFINITIONS. 

366.  A  Helated  Unit  is  a  unit  which  has  an  invariable 
relation  to  one  or  more  other  units. 

Thus,  1  foot  =  13  inches,  or  i  of  a  yard  ;  hence,  1  foot  has  an  invaria- 
ble relation  to  the  units  inch  and  yard,  and  is  therefore  a  r dated  unit. 

367.  A  Deno^ninate  dumber  is  a  concrete  number 
(15)  whose  unit  (14)  is  a  related  unit. 

Thus,  17  yards  is  a  denominate  number,  because  its  unit,  yard,  has  an 
invariable  relation  to  the  waits,  foot  and  inch,  1  yard  making  always  3  feet 
or  38  inches. 

368.  A  JDenominate  Fraction  is  a  fraction  of  a 

related  unit. 

Thus,  f  of  a  yard  is  a  denominate  fraction. 

369.  The  Orders  of  related  units  are  called  Denomi" 
nations. 

Thus,  yards,  feet,  and  inches  are  denominations  of  length ;  dollars, 
dimes,  and  cents  are  denominations  of  money. 

370.  A  Coinpoiind  Number  consists  of  several  num- 
bers expressing  related  denominations,  written  together  in  the 
order  of  the  relation  of  their  units,  and  read  as  one  number. 

Thus,  23  yd.  2  ft.  9  in.  is  a  compound  number. 

371*  A  Standard  Unit  is  a  unit  established  by  law  oc 
custom,  from  which  other  units  of  the  same  kind  are  derived. 

Thus,  the  standard  unit  of  measures  of  extension  is  the  yard.  By 
dividing  the  yard  into  three  equal  parts,  we  obtain  the  umifoot ;  into  36 
equal  parts,  we  obtain  the  unit  inch  ;  multiplying  it  by  5^,  we  obtain  the 
unit  rod,  and  so  on. 

373.  Reduction  of  Denominate  Numbers  is  the 

process  of  changing  their  denomination  without  altering  their 
value. 


TABLE 

OP 

UKTTS. 

24:  gr. 

= 

1 

pwt 

20  2)wL 

=z 

1 

OZ. 

12  oz. 

=: 

1 

Ih, 

S.2gr. 

= 

1 

carat. 

60  DENOMINATE    NU3IBERS. 


UI^ITS    OF    WEIGHT. 

374.  The  Troy  pound  of  the  mint  is  the  Standard 
Unit  of  weight. 

TROY    WEIGHT. 

1.  Denominations,  —  Grains  {gr.\ 
Pennyweights  {pwt^,  Ounces  (02.),  Pounds 
(Jb.),  and  Carats. 

3.  Equivalents.— 1  lb.  =  13  oz  =r  340 
pwt.  =  5760  gr. 

3.  Use, — Used  in  weighing  gold,  silver, 
and  precious  stones,  and  in  philosophical 
experiments. 

AVOIRDUPOIS    WEIGHT. 

TABLE  OP  UNITS.  1.    Denominations,  —  Ounces   {oz,\ 

16  oz.      =    1  lb,  pounds  (^6.),  hundredweights  (cic^.),  tons  (71). 

-^^Q  jlf       __    2   c2Vf  ^'   -KQ'^**^^^^**^*'— 1  Ton  =  20  cwt.  = 

^^      *            -,    rv   '  3000  lb.  =  33000  oz. 

20    CWi.    =    1     T.  o      xr  TT      J     •  •    I,- 

3.    iJse.  —  Used   in   weighing  groceries, 
drugs  at  wholesale,  and  all  coarse  and  heavy  articles. 

4.  In  the  United  States  Custom  House,  and  in  wholesale  transactions 
in  coal  and  iron,  1  quarter  =  38  lbs.,  1  cwt.  =  113  lb.,  1  T.  r:r  3340  lb. 
This  is  usually  called  the  Long  Ton  table. 


APOTHECARIES'    WEIGHT. 

TABLE  OF  UNITS.  1.   Denominations.  —  Grains  ^gr.), 

20  gr,  =zlsc,OV  3,  Scruples  O),    Drams  (3),   Ounces  (3), 

3  3   =ldr.ov  z.         ^^^^^^^  ^^''•)- 

o   ^    _  1  ^^  ^^  z  3-  JEquiva7ei,ts.-lb.  1  =  112  =zm 

»   3     —  IOZ.OT   t.  =3  288  =  gr.  5760. 

iZ  OZ.   =  1  to.  3    jjsf>^ — Used  in  medical  prescriptions. 

4.  Medical  prescriptions  are  usually 
written  in  Roman  notation.  The  number  is  written  after  the  symbol, 
and  the  final  "  i "  is  always  written  j.     Thus,  3  vij  is  7  ounces. 


DENOMINATE     NUMBERS. 


61 


Comparative   Table  of  Units  of  WeighU 

TROT.  AVOIRDin?OIS.  APOTHE( 

1  pound     =     5760  grains     =  7000  grains     =     5760  grains. 
1  ounce     =      480      "         =      437.5    "        =      480      " 


Table  of  Avoirdupois  Pounds  in  a  Bushel,  as  Established 
by  Law  in  the  States  named. 


Wheat 

Indian  Corn. . . 

Oat? 

Barley 

Buckwheat 

Rye 

Clover  Seed . . . 
Timothy  Seed. 


60  60 
52:56 
32  32 
48'48 
40;50 
54  56 

oieo 

45:4) 


60  60  |60 
50|56  56 
35|33J;32 
48|48 
52  52 
5656 
60|60 
45:45 


s:^!^ 


160  60 
156  56 
30  30  32 
46  48 
46  42 
56  56 
160 


60  60  60 


56:58 
30 1 32 
48:48 
50  48 
56  56 
64i60 
44 


56 


60  60  60  60 
5(5  56  5656 
34  32  32  36 
46  47  4645 
42  48  4642 
5656  56  56 

601   I   ieo 


54 


50 


Peas,  Beans,  and  Potatoes  are  usually  weighed  60  lb.  to  the  bushel. 

The  following  are  also  in  use : 

100  lb.  of  Grain  or  Flour  =  1  Cental.        200  lb.  Pork  or  Beef  =  1  Barrel. 

100  "    of  Dry  Fish  =  1  Quintal.       196  "    Flour  =  1  Barrel. 

100  "    of  Nails  =  1  Keg.  240  "    Lime  =  1  Cask. 

280  lb.  salt  at  N.  Y.  Salt  Works  -  1  Barrel. 


PROBLEMS    ON    RELATED    UOTTS. 

375.     Prob.  L — To  reduce  a  denominate  or  a  com- 
pound number  to  a  lower  denomination. 

Reduce  23  lb.  7  oz.  9  pwt.  to  pennyweights. 

2  3  lb.  7  oz.  9  pwt.        Solution. — 1.  Since  12  oz.  make  1  lb,,  in  any 
-^  2  number  of  pounds  there  are  12  times  as  many 

ounces  as  pounds.  Hence  we  multiply  the 
23  lb.  by  12,  and  add  the  7  oz.,  giving  283  oz. 
2.  Again,  since  20  pwt.  make  1  oz.,  in  any 
number  of  ounces  there  are  20  times  as  many 
pennyweights  as  ounces.    Hence  we  multiply 


2  8  3  OZ. 
20 


5  6  6  9  pwt. 

the  283  oz.  by  20,  and  add  the  9  pwt.,  giving  5669  pwt. 


6»  DE  NOMINATE    NU3IBERS. 

378.  Prob.  II. — To  reduce  a  denominate  number  to 
a  compound  or  a  higher  denominate  number. 

Eeduce  7487  so.  to  a  compound  number. 

3  )  7  4  8  7  sc.  Solution.— Since  3  sc.  make  1  dr., 

o  \n  A  qT  t       ,    n  7487  sc.  must  make  as  many  drams  as 

0  )  ±±^  ar.  -f-  ^  SO.  g  jg  contained  times  in  7487,  or  2495  dr. 

1  2 )311  oz.  +  7  dr.  +2  sc. 

2  5  lb   11  07  ^'  ^^"^®  ^  ^^'  ™^^6  1  oz.,  2495  dr. 

must  make  as  many  ounces  as  8  is  con- 
tained times  in  2495,  or  311  oz.  +  7  dr. 

3.  Since  12  oz.  make  1  lb.,  311  oz.  must  make  as  many  pounds  as  12 
is  contained  times  in  311,  or  25  lb.  +  11  oz.  Hence,  7487  sc.  are  equal  to 
the  compound  number  25  lb.  11  oz.  7  dr.  2  sc. 

381.  Prob.  III. — To  retluce  a  denominate  fraction  or 
decimal  to  integers  of  lower  denominations. 

Keduce  f  of  a  ton  to  lower  denominations. 

(1.)  f  T.  =  f  of  20  cwt.  =  f  X  20  =  14  cwt.  +  f  cwt. 
(2.)  f  cwt.  =  -I  of  100  lb.  =  f  X  100  :=  28  lb.  +  4  lb. 
(3.)     4-  lb.  =  4  of  16  oz.  =  4  X  16  =  91  oz. 

Solution. — Since  20  cwt.  is  equal  1  T.,  f  of  20  cwt.,  or  14f  cwt., 
equals  f  of  1  T.  Hence,  to  reduce  the  f  of  a  ton  to  hundredweights,  we 
take  f  of  20  cwt.,  or  multiply,  as  shown  in  (1),  the  f  by  20,  the  number 
of  hundredweights  in  a  ton. 

In  the  same  manner  we  reduce  the  f  cwt.  remaining  to  pounds,  as 
shown  in  (2),  and  the  ^  lb.  remaining  to  ounces,  as  shown  in  (3). 

384.  Prob.  IY. — To  reduce  a  denominate  fraction  or 
decimal  of  a  lower  to  a  fraction  or  decimal  of  a  higher 
denomination. 

Eeduce  f  of  a  dram  to  a  fraction  of  a  pound. 
(1.)    I  dr.     =  \oz.    X  I    =  tV  oz. 
(2.)    ^  oz.  =  ^  lb.  X  A  =  -ih  lb. 
Solution.— 1.  Since  8  drams  =  1  ounce,  1  dram  is  equal  I  of  an  oz., 
and  §  of  a  dram  is  equal  ^  of  |  oz.    Hence,  as  shown  in  (1),  f  dr.=  -^^  oz. 


DENOMINATE     NUMBERS,  63 

2.  Since  12  ounces  =  1  pound,  1  ounce  is  equal  ^^  of  a  pound,  and,  as 
shown  in  (2),  5%  of  an  ounce  is  equal  ^q  of  ^^  lb.,  or  y^^  lb.    Hence,  ?  dr. 

387.  Pkob.  V. — To  reduce  a  compound  number  to  a 
fraction  of  a  higher  denomination. 

Reduce  §4  36  32  to  a  fraction  of  1  pound. 

(1.)         !  4  3  6  33  =  3116;      lb.  1  =  3288. 

(2.)        iM  =  f I ;    hence,     1  4  3  6  32  =  lb.  ff- 

Solution. — 1.  Two  numbers  can  be  compared  only  when  they  are  the 
same  denomination.  Hence  we  reduce,  as  shown  in  (1),  the  3  4  3  6  32 
and  the  lb.  1  to  scruples,  the  lowest  denomination  mentioned  in  either 
number. 

2.  §4  36  32  being  equal  3116,  and  lb.  1  being  equal  3288,  §4  36 
32  is  the  same  part  of  lb.  1  as  3 116  is  of  3288,  which  is  |^f,  or  |f. 
Hence  §4  3  6  32  =  lb.  f  |,  or  .004027. 

15.  Reduce  8  cwt.  3  qr.  16  lb.  to  the  decimal  of  a  ton. 

a  K\-\  c   Oftlh  Abbreviated  Solution.— Since  the  16 

I '. *  pounds  are  reduced  to  a  decimal  of  a  quar- 

4)3.64  qr.  ter  by  reducing  ^  to  a  decimal,  we  annex 

o  /\  \  o    qT  __^i.  two  ciphers  to  the  16,  as  shown  in  the  mar- 

' — '- *  gin,  and  divide  by  25,  giving  .64  qr. 

.  4  4  5  5  T.  To  this  result  we  prefix  the  3  quarters, 

3  64 
giving  8.64  qr.,  which  is  equivalent  to  -^  hundredweights  ;  hence  we 

divide  by  4,  as  shown  in  the  margin,  giving  .91  cwt. 

To  the  result  we  again  prefix  the  8  c^vt.,  giving  8.91,  which  is  equiva- 

8  91 
lent  to  -'^  of  a  ton,  equal  .4455  T.    Hence,  8  cwt.  3  qr.  16  lb.  =  .4455  T. 

16.  Reduce  8  oz.  6  dr.  2  sc.  to  the  decimal  of  a  pound. 

17.  What  decimal  of  24  lb.  Troy  is  2  lb.  8  oz.  16  pwt.? 

18.  9  oz.  16  pwt.  12  gr.  are  what  decimal  of  a  pound? 

19.  Reduce  12  cwt.  2  qr.  18  lb.  to  the  decimal  of  a  ton. 

20.  What  decimal  of  a  pound  are  §  9  3  5  32  gr.  18  ? 

21.  Reduce  11  oz.  16  pwt.  20  gr.  to  the  decimal  of  a  pound. 

22.  Reduce  7  lb.  5  oz.  Avoir,  to  a  decimal  of  12  lb.  5  oz.  3  pwt.  Troy. 

23.  1  lb,  9  oz.  8  pwt.  is  what  part  of  3  lb.  Apoth,  weight  ? 


64  DENOMINATE     NUMBERS. 

390.  Prob.  VI. — To  find  the  sum  of  two  or  more  de- 
nominate or  compound  numbers,  or  of  two  or  more 
denominate  fractions. 

1.  Find  the  sum  of  7  cwt.  84  lb.  14  oz.,  5  cwt.  97  lb.  8  oz., 
and  2  cwt.  9  lb.  15  oz. 

cwt.        lb.        oz.  Solution. — 1.  We  write  numbers  of  the 

7        8  4        14  same  denomination  under  each  other,  as  shown 

c        Q  w  g  in  the  margin. 

jj  0        1^  ^'  ^®  ^^^   ^^   ^^   Simple    Numbers,  com- 

mencing  with  the  lowest  denomination.   Thus, 

15        9  2  5  15,  8,  and  14  ounces  make  37  ounces,  equal 

2  lb.  5  oz.      We  write  the  5  oz.  under  the 
ounces  and  add  the  2  lb.  to  the  pounds. 

We  proceed  in  the  same  manner  with  each  denomination  until  the 
entire  sum,  15  cwt.  92  lb.  5  oz.,  is  found. 

2.  Find  the  sum  of  ^  lb.,  f  dr.,  and  j  sc. 

Solution.— 1.  According  to  (261), 
only  fractional    units    of    the    same 
kind  and  of  the   same  whole  can  be 
added ;  hence  we  reduce  f  lb.,  f  dr., 
and  f  sc.  to  integers  of  lower  denomi- 
5        3        2        3      nations    (381),    and    then    add    the 
results,  as  shown  in  the  margin.     Or, 
2.  The  given  fractions  may  be  reduced  to  fractions  of  the  same  de- 
nomination (384),  and  the  results  added  according  to  (261),  and  the 
value  of  the  sum  expressed  in  integers  of  lower  denominations  according 
to  (381). 

393.  Prob.  VII. — To  find  the  difference  between  any 
two  denominate  or  compound  numbers,  or  denominate 
fractions. 

Find  the  difference  between  27  lb.  7  oz.  15  pwt.  and  13  lb. 
9  oz.  18  pwt. 

Solution. — 1.  We  write  numbers  of  the  same 
denomination  under  each  other. 

2.  We  subtract  as  in  simple  numbers.    W^hen 

the  number  of  any  denomination  of  the  subtra- 

13        9        17  hend  cannot  be  taken  from  the  number  of  tlie 


oz. 

dr. 

sc.      gr. 

-Jib.  =  5 

2 

2      0 

idr.  = 

2      8 

}  sc.  = 

15 

lb. 

oz. 

pwt. 

27 

7 

15 

13 

9 

18 

DENOMINATE     NUMB  ERS.  G5 

same  denomination  in  the  minuend,  we  add  as  in  simple  numbers 
(05 — III)  one  from  the  next  higher  denomination.  Thus,  18  pwt.  can- 
not be  taken  from  15  pwt. ;  we  add  1  of  the  7  oz.  to  the  15  pwt.,  making 
35  pwt.  18  pwt.  from  35  pwt.  leaves  17  pwt.,  which  we  write  under  the 
pennyweights. 

We  proceed  in  the  same  manner  with  each  denomination  until  the 
entire  difference,  13  lb.  9  oz.  17  pwt.,  is  found. 

To  subtract  denominate  fractions,  we  reduce  as  directed  in  addition, 
and  then  subtract. 

394.  Peob.  VIII. — To  multiply  a  denominate  or  com- 
pound number  by  an  abstract  number. 

Multiply  18  cwt.  74  lb.  9  oz.  by  6. 

18  cwt.     74  lb.     9  oz.  Solution.— We  multiply  as 

a  in  simple  numbers,  commencing 

with  the  lowest  denomination. 


5  T.      12  cwt.     47  lb.     6  oz.         Thus,    6    times    9    oz.    equals 

54  oz.  We  reduce  the  54  oz.  to 
pounds  (378),  equal  3  lb.  6  oz.  We  write  the  6  oz.  under  the  ounces, 
and  add  the  3  lb.  to  the  product  of  the  pounds. 

We  proceed  in  this  manner  with  each  denomination  until  the  entire 
product,  5  T.  12  cwt.  47  lb.  6  oz.,  is  found. 

396.  Prob.  IX. — To  divide  a  denominate  or  com- 
pound number  by  any  abstract  number. 

Divide  29  lb.  7  oz.  2  dr.  by  7. 

lb.      oz.     dr.  Solution.— 1.  The  object  of  the  division 

7)29        7        2  is  to  find  i  of  the  compound  number.     This 

7       ^        I  is  done  by  finding  the  \  of  each  denomina- 

tion separately.    Hence  the  process  is  the 
same  as  in  finding  one  of  the  equal  parts  of  a  concrete  number. 

Thus,  the  |  of  29  lb.  is  4  lb.  and  1  lb.  remaining.  We  write  the  4  lb. 
in  the  quotient,  and  reduce  the  1  lb.  to  ounces,  which  added  to  7  oz. 
make  19  oz.     We  now  find  the  }  of  the  19  oz.,  and  proceed  as  before. 

1.  Divide  9  T.  15  cwt.  3  qr.  18  lb.  by  2  ;  by  5 ;  by  8 ;  by  12. 

2.  If  29  lb.  7  oz.  16  pwt.  are  made  into  7  equal  parts,  how 
much  will  there  be  in  each  part  ? 


66  DENOMINATE    NUMBERS. 


TJITITS    OF    LENGTH. 

397.  A  yard  is  the  Standard  Unit  in  linear,  surface, 
and  solid  measure. 

LINEAR    MEASURE. 

TABLE  OF  UNITS.  1.  Denominations, — Inches  (in.),  Feet 

12     in,  =   1ft,  (ft.),  Yards  (yd.),  Rods  (rd.),  Miles  (mi.). 

S    ft     z=z   1  yd  ^'  Equivalents,—!  mi.  =  320  rd.  =  5280 

6'^  v'd  —  Ird         ft.  =  63360  in. 
^  "   *  *  3.  Use. — Used  in  measuring  lines  and  dis- 

320    rd.  =   1  7iii,         tances. 

4.  In  measuring  cloth,  the  yard  is  divided 
iato  Tialves,  fourtlis,  eighths,  &nd  sixteenths.  In  estimating  duties  in  the 
Custom  House,  it  is  divided  into  tenths  and  hundredths. 


Table  of  Special  Denominations, 

60      Geographic  or)       _  ^  n  3  ^^  Latitude  on  a  Meridian  or  of 

69.16  Statute  Miles   S       ~        ^^^^  \  Longitude  on  the  Equator. 
360      Degrees  =  the  Circumference  of  the  Earth. 

1.16  Statute  Miles  =  1  Geog.  Mi,  Used  to  measure  distances  at  sea. 

3      Geographical  Miles=  1  League 

^      Feet  =  1  Fathom.     Used  to  measure  depths  at  sea. 

Used  to  measure  tlie  height  of  horses 


4      Inches  =  1  Hand.  \ 


at  the  shoulder 


SURVEYORS'    LINEAR   MEASURE. 

TABLE  OF  UNITS.  1.  Denominations,— Itiiiis.  (1.),  Rod  (rd.), 

7.92  in.  =  1  Z.  Chain  (ch.).  Mile  (mi.). 

25  Z      =1  rd.  ^-  Equivalents,—!  mi. -80  ch.  =  320  rd. 

A   */7  —  1     7  *         =8000  L 

^ra,  —  I  Ctl.^  g    jj^^^^  _  jj^^^  jjj  measuring  roads  and 

80  cTl.  =  1  mi.        boundaries  of  land. 

4.  The  Unit  of  measure  is  the  Ounter's 

Chain,  vjrhich  contains  100  links,  equal  4  rods  or  66  feet. 


DENOMINATE     NUMBERS, 


67 


UE'ITS    OF    SUEFACE. 

399.  A  square  yard  is  the  Standard  Unit  of  surface 
measure. 

400.  A  Surface  has  two  dimensions — length  and  hreadth, 

401.  A  Square  is  a  jyZa/^e  surface  bounded  by  four  equal 
lines,  and  having  four  right  angles. 

402.  A  Mectangle  is  any  plane  surface  having  four  sides 
and  four  right  angles. 

403.  The  Unit  of  Measure  for  surfaces  is  usually  a 
square,  each  side  of  which  is  one  unit  of  a  known  length. 

Thus,  in  14  sq.  ft.,  tlie  unit  of  measure  is  a  square  foot. 

404.  The  Area  of  a  rectangle  is  the  surface  included 
within  its  boundaries,  and  is  expressed  by  the  number  of 
times  it  contains  a  given  U7iit  of  measure. 

Thus,  since  a  square  yard  is  a  surface,  each 
side  of  which  is  3  feet  long,  it  can  be  divided 
into  3  rows  of  square  feet,  as  shown  in  the 
diagram,  each  row  containing  3  square  feet. 
Hence,  if  1  square  foot  is  taken  as  the  Unit  of 
Measure,  the  area  of  a  square  yard  is 
3  sq.  ft.  X  3  =  9  sq.  ft. 

The  area  of  any  rectangle  is  found  in  the 
same  manner.  3  sq.  ft,  x  3  =  -9  sq.  ft. 


3  feet  long. 


SQUARE    MEASURE. 


TABLE  OF  UNITS. 


144  sq.  in. 
^  sq.ft. 
30^  6-^.  yd. 
100  sq.  rd. 
640^. 


1  sq.ft. 
1  sq.  yd. 
1  sq.  rd.,  or  P. 

1  sq.  mi. 


1.  Denortiinntioiis. — Square 
Inch  (sq.  in.).  Square  Yard  (sq.  yd.), 
Square  Rod  (sq.  rd.),  Acre  (A.), 
Square  Mile  (sq.  mi.). 

2.  Eqifivalents. — 1  sq.  mi.  = 
640  A. = 102400  sq.  rd. = 3097600  sq. 
yd.=  27878400  sq.  ft.  =  4014489000 
sq.  in. 


TABLE 

OF 

UNITS. 

625  sq. 

I. 

IP. 

16  P. 

z=z 

1  sq.  cli. 

10  sq. 

cli. 

\A. 

640  X 

;  1  sq.  mi. 

36  sq. 

mi. 

:  1  Tp. 

68  DENOMINATE     NUMBERS. 

3.  Use, — Used  in  computing  areas  or  surfaces. 

4.  Glazing  and  stone-cutting  are  estimated  by  the  square  foot ;  plaster- 
ing, paving,  painting,  etc. ,  by  the  square  foot  or  square  yard ;  roofing, 
flooring,  etc.,  generally  by  the  square  of  100  square  feet. 

5.  In  laying  shingles,  one  thousand,  averaging  4  inches  wide,  and  laid 
5  inches  to  the  weather,  are  estimated  to  cover  a  square. 


SURVEYOKS'    SQUARE    MEASURE. 

1.  Denotninatiotis.—Sqa&Te  Link 
(sq.  1.),  Poles  (P.),  Square  Chain  (sq.  ch.), 
Acres  (A.),  Square  Mile  (sq.  mi.),  Town- 
ship (Tp.). 

2.  Equivalents, — 1  Tp.=36  sq.  mi. 
=23040  A.=230400  sq.  ch.  =  3686400  P. 
=  2304000000  sq.l. 

3.  Use. — Used  in  computing  the  area 
of  land. 

4.  The  Unit  of  land  measure  is  the  acre.  The  measurement  of  a 
tract  of  land  is  usually  recorded  in  square  miles,  acres,  and  hundredths  of 
an  acre. 

insriTS    OF    VOLUME. 

408.  A  Solid  or  Volume  has  three  dimensions — length, 
hreadtli,  and  thichness. 

409.  A  Mectangular  Solid  is  a  body  bounded  by  six 
rectangles  called  faces. 

410.  A  Cube  is  a  rectangular  solid,  bounded  by  six  equal 
squares. 

411.  The  Unit  of  Measure  is  a  cube  whose  edge  is  a 
unit  of  some  known  length. 

412.  The  Volume^  or  Solid  Contents  of  a  body  is 
expressed  by  the  number  of  times  it  contains  a  given  unit  of 
measure.  For- example,  the  contents  "of  a  cubic  yard  is 
expressed  as  27  cuUcfeet. 


DENOMINATE     NUMBERS. 


G9 


Thus,  piuce  each  face  of  a 
cubic  yard  contaius  9  square  feet, 
if  a  section  1  ft.  thick  is  taken  it 
must  contain  3  times  3  cu.  ft.,  or 
9  cu.  ft. ,  as  shown  in  the  diagram. 

And  since  the  cubic  yard  is 
3  feet  thick,  it  must  contain  3  sec 
tions,  each  containing  9  cu.  ft., 
which  is  37  cu.  ft. 

Hence,  the  volume  or  contents 
of  a  cubic  yard  expressed  in  cubic  feet,  is  found  by  taking  the  product 
of  the  numbers  denoting  its  3  dimensions  in  feet. 

The  conterds  of  any  rectangular  solid  is  found  in  the  same  manner. 


^ 


CUBIC    MEASURE. 


TABLE  OF  UNITS. 

1728  CU.  in,  =  1  cu.ft. 
27  cu.  ft.  =  1  cti.  yd. 


1.  Denominations. — Cubic  Inch 
(cu.  in.),  Cubic  Foot  (cu.  ft.),  Cubic 
Yard  (cu.  yd.). 

2.  Eqnivnleiits.—l  cu.  yd.  =  27 
cu.  ft.  =  46656  cu.  in. 


3.  Use, — Used  in  computing  the  volume  or  contents  of  solids. 


Table  of  Units  for  Measuring  Wood  and  Stone. 

IG  cu.ft.  =  1  Cord  Foot  (cd.fi.)  \  Used  for  raeasur- 

8  cd.ft.  ^^\     —  I  n    .A(  fj)  \  i^g  ^^^^i 

128  cu.  ft.       )     ~  )    wood  and  stone. 

24J  cu.ft.  =  1  perch  {pcli.)  of  stone  or  masonry. 

1  cu.  yd.  of  earth  is  called  a  load. 

1.  The  materials  for  masonry  are  usually  estimated  by  the  cord  or 
percli,  the  work  by  the  perch  and  cubic  foot,  also  by  the  square  foot  and 
square  yard. 

2.  In  estimating  the  mason  work  in  a  building,  each  wall  is  measured 
on  the  outside,  and  no  allowance  is  ordinarily  made  for  doors,  windows, 
and  cornices,  unless  specified  in  contract.  In  estimating  the  material, 
the  doors,  windows,  and  cornices  are  deducted. 


70 


DENOMINATE    NU3[BERS. 


3.  Brickwork  is  usually  estimated  by  the  thousand  bricks.  Tlie  size  of 
a  brick  varies  thus :  North  River  bricks  are  8  in.  x  3|  x  2|,  Philadelphia 
and  Baltimore  bricks  are  8|  in.  x  4|  x  2|,  Milwaukee  bricks  8|  in.  x  4|  x  2|, 
and  Maine  bricks  7^  in.  x  3f  x  2|. 

4.  Excavations  and  embankments  are  estimated  by  the  cubic  yard. 


BOAKD    MEASUEE. 


TABLE  OF  UNITS. 

12  B.  in.  =  13.  ft. 
12  B.  ft.  =  1  cu.  ft. 


418.  A  Board  Foot  is  1  ft. 

long,  1  ft.  wide,  and  1  in.  thick. 

Hence,  12  board  feet  equals  1  cu.  ft. 

419.  A  Board  Inch  is  1  ft.  long,  1  in.  wide,  and  1  in. 

thick,  or  yV  ^^  ^  board  foot.    Hence,  12  board  inches  equals 

1  board  foot. 

Observe  carefully  the  following : 

1.  Diagram  1  represents  a  board 


(1) 

4  feet  long. 


Square 
loot. 

^ 

« 

4  X  2  =  8  sq.  ft.  or  8  B.  ft. 

(2.) 

4  feet  long. 


where  both  dimensions  are  feet. 
Hence  the  product  of  the  two  di- 
mensions gives  the  square  feet  in 
surface  (405),  or  the  number  of 
board  feet  when  the  lumber  is  not 
more  than  1  inch  thick. 

2.  Diagram  (2)  represents  a  board 
where  one  dimension  is  feet  and 
the  other  inches.  It  is  evident 
(418)  that  a  board  1  foot  long, 
1  inch  thick,  and  any  number  of 
inches  wide,  contains  as  many  board 
inches  as  there  are  inches  in  the 
width.  Hence  the  number  of  square 
feet  or  board  feet  in  a  board  1  inch 

thick  is  equal  to  the  length  in  feet  multiplied  by  the  width  in  inches 

divided  by  12,  the  number  of  board  inches  in  a  board  foot. 

3.  In  case  the  lumber  is  more  than  1  inch  thick,  the  number  of  board 

feet  is  equal  to  the  number  of  square  feet  in  the  surface  multiplied  by 

the  thickness. 

16.  Find  the  length  of  a  stick  of  timber  8  in.  by  10  in., 
which  will  contain  20  cu.  ft. 

Operation.— (1728  x  20)-*-(8  x  10)=432;  432-5-12  =  36  ft.,  the  length. 


1  ft.  by 
9  in. 

°*  4x9=36B.iu.;  36B.in.-i-12=3B.ft. 


TABLE  OF  UNITS. 

60  sec. 

=  1  771171. 

60  min. 

=  1  hr. 

24  hr. 

=  Ida. 

1  da. 

:=   1  ivk. 

365  da. 

=  1  common  yr. 

366  da. 

=  1  /e«;?  ?^r. 

100  yr. 

1=  1  ceil. 

DE  NO  311  NATE    NUMBERS.  71 


UNITS    OF    TIME. 

424.  The  mean  solar  day  is  the  Standard  Unit  of  time. 

1.  Denominations. — Seconds 
(sec).  Minutes  (min.),  Hours  (hi*.), 
Days  (da.),  Weeks  (wk.),  Months 
(mo.),  Years  (yr.),  Centuries  (cen.). 

3.  There  are  12  Calendar  Months 
in  a  year;  of  these,  April,  June, 
September,  and  November,  have 
30  da.  each.  All  the  other  months 
except  February  have  31  da.  each. 
February,  in  common  years,  has 
28  da.,  in  leap  years  it  has  29  da. 
3.  In  computing  interest,  80  days  are  usually  considered  one  month. 
For  business  purposes  the  day  begins  and  ends  at  12  o'clock  midnight. 

425.  The  reason  for  commo7i  and  leap  years  will  be  seen 
from  the  following : 

The  true  year  is  the  time  the  earth  takes  to  go  once  around  the  mn, 
which  is  365  days,  5  hours,  48  minutes  and  49.7  seconds.  Taking 
385  days  as  a  common  year,  the  time  lost  in  the  calendar  in  4  years  will 
lack  only  44  minutes  and  41.2  seconds  of  1  day.  Hence  we  add  1  day  to 
February  every  fourth  year,  making  the  year  366  days,  or  Ijeap  Year, 
This  correction  is  44  min.  41.2  sec.  more  than  should  be  added,  amounting 
in  100  years  to  18  hr.  37  miti.  10  sec.  ;  hence  at  the  end  of  100  years  we 
omit  adding  a  day,  thus  losing  again  5  hr.  22  min.  50  sec,  which  we  again 
correct  by  adding  a  day  at  the  end  of  400  years. 

How  many  yr.,  mo.,  da.  and  hr.  from  6  o'clock  P.  M.,  July  19, 
1862,  to  6  o'clock  A.  M.,  April  9,  1876. 

Solution. — 1.  Since  the  latter  date 

denotes  the  greater  period  of  time,  it 

is  the  minuend,  and  the  earlier  date, 

the  subtrahend. 

13  8        19        13  2.  Since  each  year  commences  with 


yr. 

mo. 

da. 

hr. 

1876 

4- 

9 

7 

1862 

7 

19 

18 

n 


DENOMINATE     NUMBERS. 


January,  and  each  day  with  12  o'clock  midnight,  7  o'clock  A.  M.,  April  9, 
1876,  is  the  7th  hour  of  the  9th  day  of  the  fourth  month  of  1876  ;  and 
6  o'clock  p.  M.,  July  19,  1862,  is  the  18th  hour  of  the  19th  day  of  the  7th 
month  of  1862.  Hence  the  minuend  and  subtrahend  are  written  as  shown 
in  the  margin. 

3.  Considering  24  hours  1  day,  30  days  1  month,  and  12  months  1  year, 
the  subtraction  is  performed  as  in  compound  numbers  (892),  and  13  yr. 
8  mo.  19  da.  13  hr.  is  the  interval  of  time  between  the  given  dates. 


CIECULAR    MEASURE. 


428.  A  Circle  is  a  plane 
figure  bounded  by  a  curved  line, 
all  points  of  which  are  equally 
distant  from  a  point  within  called 
the  centre. 

429.  A  Circumference  ia 

the  line  that  bounds  a  circle. 

430.  A  Degree  is  one  of  the 

360  equal  parts  into  which  the 
circumference  of  a  circle  is  sup- 
posed to  be  divided. 

431.  The  degree  is  the  Standard  Unit  of  circular 
measure. 


1.  Denominations.— Seconds  ("),  Minutes 
0,  Degrees  (°),  Signs  (S.),  Circle  (Cir.). 

2.  One-half  of  a  circumference,  or  180°,  as 
shown  by  the  figure  in  the  margin,  is  called  a 
Semi-circumference;  One-fourth,  or  90°,  a  Quad- 
rant; One-sixth,  or  60°,  a  Sextant;  and  One- 
twelfth,  or  30°,  a  Sign. 

3.  The  length  of  a  degree  varies  with  the  size 
of  the  circle,  as  will  be  seen  by  examining  the  foregoing  diagram. 

4.  A  degree  of  latitude  or  a  degree  of  longitude  on  the  Equator  is 
69.16  statute  miles.  A  minute  on  the  earth's  circumference  is  a  geograph- 
ical or  nautical  mile. 


TABLE  OF  UNITS. 

60"    ==  1' 
60'     =  1° 
30°    =18. 
12  /S'.  =  1  Cir. 
360°    =  1  Cir. 


DENOMINATE    NUMBERS, 


73 


SPECIAL   UNITS. 


Table  for  Paper, 

24  Sheets     =  1  Quire. 
30  Quires     =  1  Ream. 

2  Reams     =  1  Bundle. 

5  Bundles  =  1  Bale. 


Table  for  Counting, 

12  Things  =  1  Dozen  (doz.) 

12  Dozen    =  1  Gross  (gro.) 

12  Gross    =  1  Great  Gross  (G.  Gro.) 

20  Things  =  1  Score  (Sc.) 


UI^ITS     OF    MO]NET 


UNITED    STATES    MONEY. 


433.  The  dollar  is 
States  money. 

TABLE  OF  UNITS. 

10  m.  —  1  ct. 
10  ct.  = 
10  d.   = 


the    Standard  Unit  of  United 


Id. 


1.  Denominations.— Mills  (m,).  Cents  (ct.), 
Dunes  (d.),  Dollars  ($),  Eagles  (E.). 

2.  The  United  States  coin,  as  fixed  by  the 
Coinage  Acts  of  1873  and  1878,  is  as  follows; 
Goldf  the  double-eagle,  eagle,  half-eagle,  quar- 

$10         =  1  ^.        ter-eagle,    three-dollar,  and  one-dollar;  Silver, 
the  trade-dollar,  dollar,  half-dollar,  quarter-dollar, 
and  dime ;  Nickel,  the  five-cent  and  three-cent ;  Bronze,  one-cent. 

3.  Composition  of  Coins. — Gold  coin  contains  .9  pure  gold  and  .1 
silver  and  copper.  Silver  coin  contains  .9  pure  silver  and  .1  pure  copper. 
Nickel  coin  contains  .25  nickel  and  .75  copper.  Bronze  coin  contains  .95 
copper  and  .05  zinc  and  tin. 

4.  The  Trade-dottar  weighs  420  grains  and  is  designed  for  commercial 
purposes  solely.     The  silver  Dollar  weighs  412^  grains. 


CANADA    MONEY. 

434.  1.  Denominations. — Mills,  Cents,  and  Dollars.  These 
have  the  same  nominal  value  as  in  United  States  Money. 

2.  The  Coin  of  the  Dominion  of  Canada  is  as  follows :  Gold,  the 
coins  in  use  are  the  sovereign  and  half-sovereign  ;  Silver,  the  fifty-cent, 
twenty-five  cent,  ten-cent,  and  five-cent  pieces ;  Bronze,  the  one-cent 
piece. 


74  DENOMINATE    NUMBERS. 


ENGLISH    MONEY. 

435.  The  pound  sterling  is  the  Standard  Unit  of 

Enghsh  money.     It  is  equal  to  $4.8665  United  States  money. 

TABLE  OP  UNITS.  1.  Denominations, — Farthings  (far,), 

^  r^^^  =  1  ^.  Pennies  (d.),  Shillings  (s,),  Sovereign  (sov.), 

-j  2  ^       -\  g  Pound  (£),  Florin  (fi.),  Crown  (cr.). 

.     '  a  ^-    "^^^  Coins  in  general  use  in   Great 

20  5.       =  -^  Britain  are  as  follows :  Gold,  sovereign  and 

(  or  £1.  half-sovereign ;  Silver,  crown,  half-crown, 

2  5.       z=z  Ifl,  florin,  shilling,  six-penny,  and  three-penny  ; 

Kg        -—  -^  Q,f.^  Coj>l)er,  penny,  half -penny,  and  farthing. 


FRENCH    MONEY. 

436.  The  silver  franc  is  the  StaJidard  Unit  of  French 
money.     It  is  equal  to  $.193  United  States  money. 

1.  Denominations. — Millimes  (m.),  Cen- 

TABLE  OF  UNITS.       ^j^^^  ^^^^^^  Decimes  (dc).  Francs  (fr.). 

10  7n.   z=  1  ct.  2.  Equivalents,—!  fr.  =  10  dc.  =  100  ct.= 

10  ct    =   1  dc.        1000  m. 

'    10  dc.  =   1  fr.  3.  The  Coin  of  France  is  as  follows  :  Gold, 

100,  40,  20,  10,  and  5  francs;  Silver,  5,  2,  and 

1  franc,  and  50  and  25  centimes ;  Dronze,  10,  5, 2,  and  1  centime  pieces. 


GERMAN    MONEY. 

437.  The  mark  is  the  Standard  Unit  of  the  Mw  Ger- 
man  Empire.  It  is  equal  to  23.85  cents  United  States  money, 
and  is  divided  into  100  equal  parts,  one  of  which  is  called  a 
Pfenniff. 

1.  The  Coins  of  the  Neio  Empire  are  as  follows  :  Gold,  20,  10,  and 
5  marks ;  Silver ,  2  and  1  mark  ;  Nickelf  10  and  5  pfennig. 

2.  The  coins  most  frequently  referred  to  in  the  United  States  are  the 
silver  Thaler,  equal  74G  cents,  and  the  silver  Groschen,  equal  2^  cents. 


DENOMINATE     NUMBERS. 


75 


THE    METRIC     SYSTEM. 

439.  The  3Iefric  System  of  Related  Units  is  formed  according  to 
ine  decimal  scale. 

440.  The  Meter,  which  is  39.37079  inches  long,  or  nearly  one 
^■en-millionth  of  the  distance  on  the  earth's  surface  from  the  equator  to 
the  pole,  is  the  base  of  the  system. 

441.  The  I*rlmary  or  JPtincipal  Units  of  the  system  are  the 
MeteVy  the  Are  (air),  the  Stere  (stair),  the  Liter  (leeter),  and  the  Gram. 
All  other  units  are  multiples  and  sub-multiples  of  these. 

442.  The  names  of  Multiple  Units  or  higher  denominations  are 
formed  by  prefixing  to  the  names  of  the  iirimary  units  the  Greek  nu- 
merals Dekc  10),  Hecto  (100),  Kilo  (1000),  and  Myria  (10000). 

443.  The  names  of  Snh-multiple  UnitSf  or  lower  denomina- 
tions, are  formed  by  prefixing  to  the  names  of  the  primary  units  the 
Latin  numerals,  Beci  (yV),  Centi  (x^ir),  and  Milli  (ttjW- 

¥NITS    OF    LENGTH. 

444.  The  Meter  is  the  prindpal  unit  of  length. 


TABLE  OF  UNITS. 

10  Millimeters, 

mm. 

= 

1  Centimeter 

= 

.3937079  in. 

10  Centimeters, 

cm. 

— 

1  Decimeter 

= 

3.937079  in. 

10  Decimeters, 

dm. 

— 

1  Meter 

= 

39.37079  in. 

10  Meters, 

M. 

= 

1  Dekameter 

=r 

32.808992  ft. 

iO  Dekameters, 

Dm,. 

= 

1  Hectometer 

= 

19.884237  rd. 

10  Hectometers, 

Hm. 

= 

1  Kilometer 

= 

.6213824  mi. 

10  Kilometers, 

Kw.. 

= 

1  Myriameter  (Mm.)  = 

6.213824  mi. 

The  meter  \B  used 

in  place 

of  one  yard  in  measuring 

cloth  and  short  distances. 

Long  distances  are  usually  measured  by  the  kilometer. 

UNITS    OF    SITEFACE. 
445*  The  Square  Meter  is  the  principal  unit  of  surfaces. 

TABLE  OF  UNITS. 

100  Sq.  Millimeters,  sq.  mm.  =  1  Sq.  Centimeter  =  .155+  sq.  in. 

100  Sq.  Centimeters,  sq.  cm.     =  1  Sq.  Decimeter  =  15.5  +  sq.  in. 

100  Sq.  Decimeters,  sq.  dm.    --==  1  Sq,  Meter  {Sq.  M.)  =  1.193+  sq.yd 


76  DENOMINATE     NUMBERS, 

446.  The  Are,  a  square  whose  side  is  10  meters,  is  the  principal 
unit  for  measuring  land. 

TABLE  OF  UNITS. 

100  Centiares,  ca.  —  1  Are  =  119.6034  sq.  yd. 

100  Ares,  A.  =  \  Hectare  {Ha.)  =      2.47114  acres. 

UNITS    OF    VOLUME. 

44*7.  The  Cubic  Meter  is  the  principal  unit  for  measuring  ordi- 
nary solids,  as  embankments,  etc. 

TABLE  OF  UNITS. 

1000  Cu.  Millimeters,  cu.  mm.  =  1  Cu.  Centimeter  =  ,061  cu.  in. 
1000  Cu  Centimeters,  cu.  cm.  =  1  Cu.  Decimeter  =  61.026  cu.  in. 
1000  Cu.  Decimeters,   cu.  dm.     =    1  Cu.  dieter       =   35.316  cu.  ft. 

448.  The  Stere,  or  Cu.  Meter,  is  the  principal  unit  for  measuring 
wood. 

TABLE  OF  UNITS. 

10  Decisteres,  (?5^.  =  1  Stere  —  35.316+  cu.  ft. 

10  Steres,      St.    =  1  Dekastere  (Z)«^.)  =  13.079+  cu.  yd. 

UNITS    OF    CAPACITY. 

449.  The  Liter  is  the  principal  unit  both  of  Liquid  and  Dry 
Measure.  It  is  equal  to  a  vessel  whose  volume  is  equal  to  a  cube  whose 
edge  is  one4enth  of  a  meter, 

TABLE  OF  UNITS. 


10  Milliliters,  ml.  =  1  Centiliter 

=      .6102  cu.  in. 

=        .338  fl.  oz. 

10  Centiliters,  d.    =  1  Deciliter 

=     6.1022  "    " 

=        .845  gill. 

10  Deciliters,   dl.    =  1  Liter 

=       .908  qt. 

=       1.0567  qt. 

10  Liters,       L.    =  1  Dekaliter 

=     9.08      •* 

=       2.6417  gal. 

10  Dekaliters,  JDl.  =  1  Hectoliter 

=    2.8372+  bu. 

=    26.417      " 

10  Hectoliters,^;.  =:  1  Kiloliter 

=  28.372+     " 

=  264.17        " 

10  Kiloliters,   Kl.  =  1  Myrialiter 

=283.72+       " 

=2641.7 

The  Hectoliter  is  used  in  measuring  large  quantities  in  both  liquid  and  dry  measure. 

UNITS    OF    WEIGHT. 

450.  The  Gram  is  the  principal  unit  of  weight,  and  is  equal  to 
the  weight  of  a  cube  of  distilled  water  whose  edge  is  one  centimeter. 


DENOMINATE     NUMBERS, 


77 


TABLE  OF  UNITS. 


10  Milligrams, 
10  Centigrams, 
10  Decigrams, 
10  Grams, 
10  Dekagrams, 


mg. 

eg. 

dg. 

a. 

Dg. 


-10  Hectograms,  Hg, 

10  Kilograms,     Kg. 
10  Myriagrams,  Mg. 

10  Quintals, 


=     1 


Centigram 

Decigram 

Gram 

Dekagram 

Hectogram 
\  Kilogram  ) 
}    or  Kilo.    S 

Myriagram 

Quintal 
Tonneau 
or  Ton. 


.15432+  gr.  Troy. 
1.54334+   "      " 
15.43248+   "      " 

.3527  +  oz.  Avoir. 
3.52739+   "      " 

2.20462+  lb. 


=   22.04621  + 
=   220.46212  + 

=    2204.6212  + 


The  Kilogram  or  Kilo.,  which  is  little  more  than  2|  lb.  Avoir.,  is  the 
common  weight  in  trade.  Heavy  articles  are  weighed  by  the  Ton- 
neau, which  is  204  lb.  more  than  a  common  ton. 


Comparative  Table  of  Units, 


1  Inch        =     .0254  meter. 

1  Cu.  foot 

= 

.2832  Hectoliter. 

1  Foot        =     .3048      " 

1  Cu.  yard 

= 

.7646  Steres. 

1  Yard        =     .9144       " 

1  Cord 

= 

3.625  Steres. 

1  Mile        =  1.6093  Kilometers. 

1  Fl.  ounce 

=: 

.02958  Liter. 

1  Sq.  inch  =     .0006452  sq.  meter. 

1  Gallon 

= 

3.786  Liters. 

1  Sq.  foot  =     .0929 

1  Bushel 

= 

.3524  Hectoliter. 

1  Sq.  yard  =     .8361 

1  Troy  grain 

= 

.0648  Gram. 

1  Acre        =40.47  Ares. 

1  Troy  lb. 

= 

.373  Kilogram. 

1  Sq.  mile  =     .259  Hectares. 

1  Avoir,  lb. 

=: 

.4536  Kilogram. 

1  Cu,  inch  =     .01639  Liter. 

1  Ton 

= 

.9071  Touneau. 

EXAMPLES     FOR      PRACTICE 


451.  Reduce 

1.  84  lb.  Avoir,  to  kilograms. 

2.  37  T.  to  tonneau. 

3.  96  bu.  to  hectoliters. 

4.  75  fl.  oz.  to  liters. 

5.  89  cu.  yd.  to  steres. 

6.  328  acres  to  ares. 

13.  If  the  price  per  gram  is  $. 


7.  4.0975  liters  to  cu.  in. 

8.  31.7718  sq.  meters  to  sq.  yd. 

9.  272.592  liters  to  bushels. 

10.  35.808  kilograms  to  Troy  gr. 

11.  133.75  steres  to  cords. 

12.  33.307  steres  to  cu.  ft. 
what  is  it  per  grain  ? 


78  DEN0  3IINATE     NUMBERS, 

14.  If  the  price  per  liter  is  $1.50,  what  is  it  per  quart? 

15.  At  26.33  cents  per  hectoliter,  what  will  be  the  cost  of  157  bushels 
of  peas  ? 

16.  When  sugar  is  selling  at  2.168  cents  per  kilogram,  what  will  be 
the  cost  of  138  lb.  at  the  same  rate? 

17.  Reduce  834  grams  to  decigrams  ;  to  dekagrams. 

18.  In  84  hectoliters  how  many  liters  ?  how  many  centiliters  ? 

19.  A  man  travels  at  the  rate  of  28.279  kilometers  a  day.     How  many 
miles  at  the  same  rate  will  he  travel  in  45  days  ? 

20.  If  hay  is  sold  at  $18,142  per  ton,  what  is  the  cost  of  48  tonneau  at 
the  same  rate  ? 

21.  When  a  kilogram  of  coffee  costs  $1.1023,  what  is  the  cost  of 
148  lb.  at  the  same  rate  ? 


EEVIEW    AND    TEST    QUESTIONS. 

4:65.  1.  Define  Related  Unit,  Denominate  Number,  Denominate 
Fraction,  Denomination,  and  Compound  Number. 

2.  Repeat  Troy  Weight  and  Avoirdupois  Weight. 

3.  Reduce  9  bu.  3  pk.  5  qt.  to  quarts,  and  give  a  reason  for  each  step 
in  the  process. 

4.  In  9  rd.  5  yd.  2  ft.  how  many  inches,  and  why  ? 

5.  Repeat  Square  Measure  and  Surveyors'  Linear  Measure. 

6.  Reduce  2345G  sq.  in.  to  a  compound  number,  and  give  a  reason  for 
each  step  in  the  process. 

7.  Define  a  cube,  a  rectangular  volume,  and  a  cord  foot. 

8.  Show  by  a  diagram  that  the  contents  of  a  rectangle  is  found  by 
multiplying  together  its  two  dimensions. 

9.  Define  a  Board  Foot,  a  Board  Inch  ;  and  show  by  diagrams  that 
there  are  12  hoard  feet  in  1  cubic  foot  and  12  hoard  inches  in  1  board  foot. 

10.  Reduce  f  of  an  inch  to  a  decimal  of  a  foot,  and  give  a  reason  for 
each  step  in  the  process. 

11.  How  can  a  pound  Troy  and  a  pound  Avoirdupois  be  compared? 

12.  Reduce  .84  of  an  oz.  Troy  to  a  decimal  of  an  ounce  Avoirdupois, 
and  give  reason  for  each  step  in  the  process. 

13.  Explain  how  a  compomid  number  is  reduced  to  a  fraction  or  deci- 
mal of  a  higher  denomination.  Illustrate  the  abbreviated  method,  and 
give  a  reason  for  each  step  in  the  process. 


PART    SECOND. 


BUSINESS   ARITHME 


imh 


SHORT    METHODS. 

466.  Practical  devices  for  reaching  results  rapidly  are  of 
first  importance  in  all  business  calculations.  Hence  the  fol- 
lowing summary  of  short  methods  should  be  thoroughly 
mastered  and  applied  in  all  future  work.  The  exercises  under 
each  problem  are  designed  simply  to  illustrate  the  application 
of  the  contraction. 

When  the  directions  given  to  perform  the  work  are  not 
clearly  understood,  the  references  to  former  explanations 
should  be  carefully  examined. 

461'.   Prob.  I.— To  multiply  by  lO,  100, 1000,  etc. 

Move  tJie  decimal  ^mnt  in  the  multiplicand  as  many  places 
to  the  right  as  there  are  ciphers  in  the  multiplier,  annexing 
ciphers  ivhen  necessary  (91). 

Multiply  the  following: 

1.  84  X 100.  4.  3.8097  x  10000.  7.  3426  x  1000. 

2.  76  X 1000.  5.  .89752  x  1000.  8.  7200  x  100000. 

3.  5. 73  X 100.  6.  3.0084  x  10000.  9.  463  x  1000000. 

468.  Prob.  II. — To  multiply  where  there  are  ciphers 
at  the  right  of  the  multiplier. 

Move  the  decimal  point  in  the  multiplicand  as  many  places 
to  the  right  as  there  are  ciphers  at  the  right  of  the  multiplier, 
annexing  ciphers  when  necessary,  and  multiply  the  result  by 
the  significant  figures  in  the  multiplier  (93). 

(207) 


80  BUSINESS    ARITHMETIC. 

Multiply  the  following : 

1.  376  X  800.             4.  836.9  x  2000.  7.  3800  x  7200. 

2.  42.9  X  420.            5.  7.648  x  3200.  8.  460  x  900. 

3.  500  X  700.             6.  2300  x  5000.  9.  .8725  x  3600. 

469.  Prob.  III.~To  multiply  by  9,  99,  999,  etc. 

Move  the  decimal  point  in  the  imdtiplicand  as  many  places 
to  the  right  as  there  are  nines  in  the  multiplier,  annexing 
ciphers  when  necessary,  and  subtract  the  given  multiplicand 
from  the  result. 

Observe  that  by  moving  the  decimal  point  as  directed,  we  multiply  by  a  number  1 
greater  than  the  given  multiplier ;  hence  the  multiplicand  is  subtracted  from  the  result. 
To  multiply  by  8,  98,  998,  and  so  on,  we  move  the  decimal  point  in  the  same  manner; 
and  subtract  from  the  result  twice  the  multiplicand. 

Perform  the  following  multiplication : 

1.  736458  X  9.  4.  53648  x  990.  7.  7364  x  998. 

2.  3895  X  99.  5.  83960  x  9999.  8.  6283  x  9990. 

3.  87634  X  999.  6.  26384  x  98.  9.  4397  x  998. 

470.  Prob.  IV.— To  divide  by  lO,  lOO,  lOOO,  etc. 

Move  the  decimal  point  in  the  dividend  as  many  places  to  the 
left  as  there  are  ciphers  in  the  divisor,  prefixing  ciphers  when 
necessary. 

Perform  the  division  in  the  following : 

1.  8736-r-lOO.  4.  23.97-j-lOOO.  7.  .54-^100. 

2.  437.2-r-lO.  5.  5.236-^100.  8.  .07-7-1000. 

3.  790.3-f-lOO.  6.  .6934-T-lOOO.  9.  7.2-f-lOOO. 

471.  Prob.  V. — To  divide  where  there  are  ciphers  at 
the  right  of  the  divisor. 

Move  the  decimal  point  in  the  dividend  as  many  places  to  the 
left  as  there  are  ciphers  at  the  right  of  the  divisor,  prefixing 
ciphers  when  necessary  (140),  and  divide  the  result  ly  the 
significant  figures  in  the  divisor  (143). 

(208) 


SHORT    METHODS,  81 

Perform  the  division  in  the  following : 

1.  7352-r-40.  4.  5.2-r-400.  7.  364.2-^640. 

2.  523.7-r-80.  5.  .96-r-120.  8.  973.5—360. 

3.  329.5-^3000.  6.  .08^-200.  9.  8.357-^600. 

472.  Peob.  YI. — To  multiply  one  fraction  by  another. 

Cancel  all  factors  common  to  a  numerator  and  a  denomina- 
pr  before  multiplying  (185 — II). 

Perform  the  following  multiplications  by  canceling  common 
factors : 

l-Hxff.  6.  ixMx«.  11.  Mx-AVxf. 

2.  If  x^V  7.  Axlf  xA.  12.  MMxtV^x^^. 

3.  tttxff.  8.  AxHxff.  13.  «xi|ix^V 

4.  ^xA-  9.  fxffxA.  14.  HfxttxJ. 

5.  exT%.  10.  «fxffXT|^.  15.  il^xyij^xf 

473.  Prob.  YII. — ^To  divide  one  fraction  by  another. 

Cancel  all  factors  common  to  both  numerators  or  common  to 
hotli  denominators  before  dividing  (291).     Or, 
Invert  tlie  divisor  and  cancel  as  directed  in  Prob.  VL 

Perform  the  division  in  the  following,  canceling  as  directed : 

1.  H-f  5.  3*A-«.  9.  i^f-Jf 

2.  H-^A-  6.  H-if.  10.  AV-tW«. 

3.  H  -^  H-  7.  AV  --  ^.  11.  .39  ^  .003. 

4.  .9  -T-  .03.  8.  .28  -i-  .04.  12.  .63  —  .0027. 

474.  Prob.  VIII. — To  divide  one  number  by  another. 

Cancel  the  factors  that  are  common  to  the  dividend  and  divi- 
sor before  dividing  (185 — II). 

Perform  the  following  divisions,  canceling  as  directed : 

1.  8400-J-300.  4.  62500-^2500.  7.  9999-^63. 

2.  3900-7-130.  5.  3420-r-5400.  8.  32000-r-400. 

3.  4635—45.  6.  89600-^800.  9.  75000—1500. 

^209^ 


BUSINESS    ARITirilETIO, 


ALIQUOT    PARTS. 

475.  An  Aliquot  Part  of  a  number  is  any  number, 
integral  or  mixed,  which  will  exactly  divide  it. 

Thus,  2,  2^,  3^,  are  aliquot  parts  of  10. 

476.  The  aliquot  parts  of  any  number  are  found  by  divid- 
ing by  2,  3,  4,  5,  and  so  on,  up  to  1  less  than  the  given  number. 

Thus,  100-^2  =  50 ;  100-^3  =  33| ;  100  -^  4  =  25.  Each 
of  the  quotients  50,  33-J,  and  25,  is  an  aliquot  part  of  100. 

477.  The  character  @  is  followed  by  the  price  of  a  unit  or 
one  article.  Thus,  7  cords  of  wood  @  14.50  means  7  cords  of 
wood  at  $4.50  a  cord. 

478.  Memorize  the  following  aliquot  parts  of  100,  1000, 

and  $1. 

Table  of  Aliquot  Parts, 


50    =  1' 

500    =  J' 

50    ct.  =  i 

33i=  i 

333|  =  I 

33|  ct.  =  i 

25    =  J 

250    =  i 

25    ct.  =  i 

20    =  i 

200    =  i 

20    ct.  =  -J- 

16|=  i 

^  of  100.    166f  =  i 

►  of  1000.  16fct.  =  i 

l^=  \ 

142f  =  1 

14f  ct.  =  1 

12i=i 

125    =  i 

12ict.  =:  i 

1U=  i 

IIH  =  i 

Hi  ct.  =  i 

10  =M 

100    =^ 

10    ct.  =-^ 

of  $1. 


479.  Peob.  IX. — To  multiply  by  using  aliquot  parts. 

1.  Multiply  459  by  33|. 

3  )  45900  Explanation. — We  multiply  by  100  by  annexing  two 

ciphers  to  tlie  multiplicand,  or  by  moving  the  decimal 

15300  point  two  places  to  the  right.  But  100  being  equal  to 
8  times  the  multiplier  33|^,  the  product  45900  is  3  times  as  large  as  tho 
required  product ;  hence  we  divide  by  3. 

(210) 


SHORT    METHODS,  83 

Perform  the  following  multiplications  by  aliquot  parts. 

2.  974  X  50.  5.  234  x  333J.  8.  4.38  x  3^. 

3.  35.8  X  16f.  6.  869  x  llj.  9.  7.63  x  142f. 
4.895x125.             7.    72xlllf.              10.58.9x250. 

Solve  the  following  examples  orally,  by  aliquot  parts. 

11.  What  cost  48  lb.  butter  @  25  ct.  ?  @  50  ct.  ?  @  33 J  ct.  ? 
Soi.UTiON. — At  $1  a  pound,  48  would  cost  $48.    Hence  at  33|  cts.  a 

pound,  which  is  ^  of  $1,  48  pounds  would  cost  i  of  $48,  which  is  $16. 

12.  What  cost  96  lb.  sugar  @  12^  ct.  ?  @  14f  ct.  ?  @  16|  ct.  ? 

13.  What  is  the  cost  of  24  bushels  wheat  @  $1.33 J? 

Solution-.— At  $1  a  bushel,  24  bushels  cost  $24 ;  at  33|  ct.,  which  is 
I-  of  $1  a  bushel,  24  bushels  cost  $8.  Hence  at  $1.33  J-  a  bushel» 
24  bushels  cost  the  sum  of  $24  and  $8,  which  is  $32. 

14.  What  cost  42  yards  cloth  @  |1.16f  ?  @  $2.14f  ? 

15.  What  cost  72  cords  of  wood  @  U.l^  ?    @  $3.25  ? 

Find  the  cost  of  the  following,  using  aliquot  parts  for  the 
cents  in  the  price. 

16.  834  bu.  wheat  @  I1.33J ;  @  $1.50 ;  @  $1.25 ;  at  $1.1 6f. 

17.  100  tons  coal  @  $4.25  ;    at  $5.50;  @  $6.12^:    @  $5.33J. 

18.  280  yd.  cloth  @$2.14f ;   @  $1.12|-;  @$3.25;   @  $2.50. 

19.  150 bbl. apples®  $4.20;  @$4.50;  ©$4.33^. 

20.  2940  bu.  oats  @  33  ct. ;  @  50  ct. ;  @  25  ct. 

21.  896  lb.  sugar  @l^•,   @  14f ;  @  16f. 

22.  What  is  the  cost  of  2960  yd.  cloth  at  37^  ct.  a  yard  ? 

25  =i  of  100,  hence  4 )  2960  ExPLANATioN.-At  $1  a  yard, 

^ 2960  yd.  will  cost  $2960.    But 

l?i=i  of  ^5,  hence     2  )    740  25  ct.  is  i  of  $1, hence  i  of  $2960 

37^                                              370  which  is  $740,  is  the  cost  at 

$1110    ^^  ^*'  ^  ^'^• 

2.  Again,  12^-  ct.  is  the  ^  of 
35  ct.,  hence  $740,  the  cost  at  25  cts.,  divided  by  2,  gives  the  cost  at 
12i  ct.,  which  is  $370.  But  25  ct.  +  12^  ct.  =  37^  hence  $740  +  $370 
or  $1110  is  the  cost  at  37^  ct. 

(211) 


84     .  BUSINGS    ARITHMETIC. 

23.  495  bu.  barley  @  75  ct. ;  @  62^  ct. ;  @  87^  ct 

24.  680  lb.  coffee  @  37^  ct. ;  @  75  ct. ;  @  60  ct. 

25.  4384  yd.  cloth  @  12|  ct. ;  @  15  ct. ;  @  30  ct. ;  @  35  ct. 
Observe,  that  10  ct.  =  yV  of  100  ct.,  and  5  ct.  =  ^  of  10  ct. 

26.  870  lb.  tea  @  60  ct ;  @  62^  ct. ;  @  80  ct. ;  @  87^  ct. 

480,  Prob.  X. — To  divide  by  using  aliquot  parts. 

1.  Divide  7258  by  33^. 

72.58  Explanation.— 1.  We  divide  by  100  by  moving  the 

g  decimal  point  two  places  to  the  left. 

c)..  y  r/A  2.  Since  100  is  3  times  33^,  the  given  divisor,  the 

quotient  72.58  is  only  ^  of  the  required  quotient ;  hence 
we  multiply  the  72.58  by  3,  giving  217.74,  the  required  quotient. 

Perform  by  aliquot  parts  the  division  in  the  following : 

2.  8730-^3J.  5.  379.6-^-33^.  8.  460.85-^250. 


3.  9764-^5. 

6.  98.54-^50. 

9.  90.638-T-25. 

4.  8.375-r-16|. 

7.  394.8 -j- 125. 

10.  73096-333^. 

Solve  the  following  examples  orally,  using  aliquot  parts. 

11.  At  33 J  ct.,  how  many  yards  of  cloth  can  be  bought 
for  $4  ? 

Solution. — Since  $1,  or  100  ct.,  is  3  times  33|  ct.,  we  can  buy  3  yards 
for  $1.    Hence  for  $4  dollars  we  can  buy  4  times  3  yd.,  which  is  12  yd. 

Observe,  that  in  this  solution  we  divide  by  100  and  multiply  by  3,  the  number  of  times 
33J,  the  given  price,  is  contained  in  100.  Thus,  $4=400  ct.,  400-1-100=4,  and  4  x  3=12. 
In  the  solution,  the  reduction  of  the  $4  to  cents  is  omitted,  as  we  recognize  at  sight 
that  100  ct,  or  $1,  is  contained  4  times  in  $4. 

12.  How  many  yards  of  cloth  can  be  bought  for  $8  @  12|^  ct.  ? 
@  14f  ct.  ?  @  33i  ct.  ?  @  16f  ct.  ?  @  25  ct.  ?  @  10  ct.  ? 
@  50  ct.  ?    @  8  ct.  ?    @  5  ct.  ?    @  4  ct.  ? 

13.  How  many  pounds  of  butter  @  33-^  ct.  can  be  bought 
for  $7?    For  $10?    For  $40? 

14.  How  much  sugar  can  be  bought  at  12|-  ct.  per  pound 
for  $3?    For  $8?    For  $12?    For  $30?    For  $120? 

(212) 


BUSINESS    PROBLEMS.  85 

Solve  the  following,  performing  the  division  by  aliquot 
parts : 

15.  How  many  acres  of  land  can  be  bought  for  $8954  at  f  25 
per  acre?  At  $50  ?  AtlSS^?  At  $125?  At$16f?  At  1250? 

16.  How  many  bushels  of  wheat  can  be  bought  for  IG354 
at  11.25  per  bushel  ?    At  $2.50  ? 

Observe,  $1.25  =  J  of  $10  and  $2.50  =  i  of  $10.  Hence  by  moving  the 
decimal  point  one  place  to  the  left,  which  will  give  the  number  of  bu. 
at  $10,  and  multiplying  by  8,  will  give  the  number  of  bu.  at  $1.35. 
Multiplying  by  4  will  give  the  number  at  $2.50. 

17.  How  many  yards  of  cloth  can  be  bought  for  $2642  at 
33i  ct.  per  yard?  At  14f  ct?  At  25  ct.?  At  $3.33^?  At 
$2.50?    At$l.lH?    At$1.42f? 

18.  What  is  the  cost  of  138  tons  of  hay  at  $12J  ?  At  14f  ? 
At  16|  ?  At  $25  ?    At  $13.50  ?    At  $15.33J  ?    At  $17.25  ? 


BUSINESS    PEOBLEMS, 
DEFINITIONS. 

481.  Quantity  is  the  amount  of  any  thing  considered  in 
a  business  transaction. 

483.  JPrice^  or  Rate,  is  the  value  in  money  allowed  for 
a  given  unit,  a  given  number  of  units,  or  a  given  part  of  a 
quantity. 

Thus,  in  74  bu.  of  wheat  at  $2  per  bushel,  the  price  is  the  value  of  a 
unit  of  the  quantity ;  in  8735  feet  of  boards  at  45  ct.  per  100  feet,  the 
price  is  the  value  of  100  units. 

483.  When  the  rate  is  the  value  of  a  given  number  of 
units,  it  may  be  expressed  as  a  fraction  or  decimal. 

Thus,  cloth  at  $3  for  4  yards  may  be  expressed  as  $5  per  yard  ;  7  for 
every  100  in  a  given  number  may  be  expressed  j^^j^  or  .07.  Hence,  f  of 
64  means  5  for  every  8  in  64  or  5  per  8  of  64,  and  .08  means  8  per  100, 

(313) 


86  BUSINESS    ARITHMETIC. 

484.  Cost  is  the  value  in  money  allowed  for  an  entire 
quantity. 

Thus,  in  5  barrels  of  apples  at  $4  per  barrel,  $4  is  tbe  price,  and  $4x5 
or  .^20,  the  entire  value  of  the  5  barrels,  is  the  cost. 

485.  JPer  Cent  means  Per  Hundred. 

Thus,  8  per  cent  of  $600  means  $8  out  of  every  $100,  which  is  $48. 
Hence  a  given  per  cent  is  the  price  or  rate  per  100. 

486.  The  Sign  of  Per  Cent  is  %.    Thus,  %%  is  read, 
8  2)er  cent. 

Since  per  cent  means  per  hundred,  any<given^e7'  cent  may  be  expressed 
with  the  sign  %  or  in  the  form  of  a  decimal  or  common  fraction ;  thus, 
1  per  cent  is  written    1%     or    .01    or    y^. 


7  per  cent  " 

« 

7% 

"     .07     *• 

T^n. 

100  per  cent  " 

<( 

100% 

"  1.00     •' 

m- 

135  per  cent  " 

t( 

135% 

"  1.35     " 

m- 

1  per  cent  " 

« 

i% 

"  m  " 

i  _ 

100 

.005. 

48T.  Percentage  is  a  certain  number  of  hundredths  of 
a  given  quantity. 

488.  Profit  and  Loss  are  commercial  terms  used  to 
express  the  gain  or  loss  in  business  transactions. 

489.  The  Profit  or  Gain  is  the  amount  realized  on 
business  transactions  in  addition  to  the  amount  invested. 

Thus,  a  man  bought  a  farm  for  $8500  and  sold  it  for  $9200.  The 
$8500  paid  for  the  farm  is  the  amount  invested,  and  the  $9200  is  the 
whole  sum  realized  on  the  transaction,  which  is  $700  more  than  what 
was  invested  ;  hence  the  $700  is  the  profit  or  gain  on  the  transaction. 

490.  The  IjOSS  is  the  amount  which  the  whole  sum 
realized  on  business  transactions  is  less  than  the  amount 
invested. 

Thus,  if  a  horse  is  bought  for  $270  and  sold  again  for  $170,  there  is  a 
loss  of  $100  on  the  transaction. 

491.  The  Gain  and  the  Loss  are  usually  expressed  as  a 
per  cent  of  the  amount  invested. 

(214) 


BUSINESS     PROBLEMS,  87 


ORAii    e:xercises. 

493.  Express  the  following  decimally: 


1.  h%. 

6. 

2.  7^. 

6. 

3.  1^%. 

7. 

4.  25%. 

8. 

9. 

207%. 

13. 

» 

10. 

125}%. 

14. 

n%' 

11. 

312|%. 

15. 

Hfc 

12. 

i%' 

16. 

-h%' 

By  135%? 

By|%? 

112%. 

We- 
ll. What  is  meant  by  8%  ? 

18.  What  is  the  difference  in  the  meaning  of  5  per  cent  and 
6  per  seven  ? 

19.  How  is  ^  per  eight  expressed  with  figures?  7  per  five? 
13  per  twenty  9    9  per  four  9 

20.  What  does  y^  mean,  according  to  (483)  ?  What  does 
f  mean,  according  to  the  same  Art.  ? 

21.  What  is  the  difference  in  the  meaning  of  f%  and 
}  of  100  ? 

22.  What  is  the  meaning  of  .OOf  ?     Of.07f?     Of  .32-^  ? 

23.  Express  .00^  with  the  sign  %  and  fractionally. 

24.  Write  in  figures  tliree  per  cent,  and  nine  per  cent 

Express  the  following  as  a  per  cent : 

25.  f                28.  143.               31.  IJ.  34.  100. 

26.  9f              29.  236.              32.  1.  35.  700. 

27.  3f.             30.  1074.            33.  3.  36.  205. 

493.  In  the  following  problems,  some  already  given  are 
repeated.  This  is  done  first,  for  review,  and  second,  to  give 
in  a  connected  form  the  general  problems  that  are  of  con- 
stant recurrence  in  actual  business.  Each  problem  should  be 
fixed  firmly  in  the  memory,  and  the  solution  clearly  under- 
stood. 

It  will  be  observed  that  Problems  VIII,  IX,  X,  and  XI, 
are  the  same  as  are  usually  given  under  the  head  of 
JPercentage.  They  are  presented  in  a  general  form,  as  the 
solution  is  the  same  whether  hundredths,  or  some  other /r«c- 
tional  parts  are  used. 

(215) 


88  BU8INBSS    ARITHMETIC. 


PROBLEMS. 

494.  Prob.  I. — To  lincl  the  cost  when  the  number  of 
units  and  the  price  of  one  unit  are  given, 

1.  What  is  the  cost  of  35  lb.  tea  @  $4? 

Solution.— Since  1  lb.  cost  $f ,  35  lb.  will  cost  35  times  $f ,  which  is 
(271)  $25. 

Find  the  cost  and  explain  the  following  orally. 

2.  64  bu.  apples  @  l|.  6.  9  boxes  oranges  (^  $4f . 

3.  24  yd.  cloth  @  $2f.  7.  18  tons  coal  @  |6f 

4.  6|  yd.  cloth  @  8|.  8.  96  cords  wood  @  %4t^. 

5.  U\  lb.  butter  @  $f  9.  8^  yd.  cloth  @  I^V 

Find  the  cost  of  the  following,  and  express  the  answer  in 
dollars  and  cents  and  fractions  of  a  cent. 

10.  84  bu.  oats  @  $|.  14.  25f  cords  wood  @  $5^. 

11.  18  bbls.  apples  @  Mf         15.  63^  Ih.  butter  @  1^. 

12.  52  yd.  cloth  @  |2f  16.  169  acr.  land  @  $27|. 

13.  83  lb.  coffee  @  $f .  17.  32f^  lb.  sugar  @  8^. 

18.  How  much  will  a  man  earn  in  19f  days  at  $2f  per  day  ? 

19.  Sold  Wm.  Henry  2b\  lb.  butter  @  28}  ct.,  17^  lb. 
coffee  @  $.33^,  and  39|f  lb.  sugar  @  I.Uf.  How  much  was 
his  bill  ? 

20.  A  builder  has  17  carpenters  employed  @  $2.25  per  day. 
How  much  does  their  wages  amount  to  for  24|  days  ? 

495.  Prob.  II. — To  find  the  price  per  unit,  when  the 
cost  and  number  of  units  are  given. 

1.  If  9  yards  cost  $10.80,  what  is  the  price  per  yard  ? 
Solution. — Since  9  yards  cost  $10.80, 1  yard  will  cost  \  of  it,  or 

$10.80  -^  9  =  $1.20.    Hence,  1  yard  cost  $1.20. 

Solve  and  explain  the  following  orally. 

2.  If  9  lb.  sugar  cost  $1.08,  what  is  the  price  per  pound  ? 

3.  At  $4.80  for  8  yards  of  cloth,  what  is  the  price  per  yard? 

(216) 


BUSINESS    PROBLEMS.  89 

4.  If  12  lb.  of  butter  cost  $3.84,  how  much  is  it  a  pound  ? 

5.  Paid  $3.42  for  9  lb.  of  coffee.  How  much  did  I  pay  per 
pound  ? 

Solve  and  explain  the  following . 

G.  Bought  236  bu.  oats  for  $90.80.    What  did  I  pay  a  bu.  ? 

7.  A  piece  of  cloth  containing  348  yd.  was  bought  for 
$515.91.     What  did  it  cost  per  yard  ?  A?is.  $1.4825. 

8.  A  farm  containing  282  acres  of  land  was  sold  for  $22184. 
What  was  the  rate  per  acre  ?  Ans.  $78.6G+. 

9.  If  85  cords  of  stone  cost  $371,875,  what  is  the  price  per 
cord?  Ans.  $4,375. 

10.  There  were  25  mechanics  employed  on  a  building,  each 
receiving  the  same  wages ;  at  the  end  of  28  days  they  were 
paid  in  the  aggregate  $1925.     What  was  their  daily  wages  ? 

11.  A  merchant  bought  42  firkins  of  butter,  each  contain- 
ing 634  lb.,  for  $735.67.    What  did  he  pay  per  pound  ? 

12.  A  farmer  sold  70000  lb.  of  hay  for  $542.50.  How  much 
did  he  receive  per  ton  ?  Ans,  $15.50. 

496.  Prob.  III.— To  find  the  cost  when  the  number  of 
units  and  tlie  price  of  any  multiple  or  part  of  one  unit 
is  given. 

1.  What  IS  the  cost  of  21  lb.  sugar  at  15  ct.  for  J  lb.  ? 

Solution,— Since  J  lb.  cost  15  ct.,  21  lb.  must  cost  as  many  times 
15  ct.  as  I  lb.  is  contained  times  in  it.  Hence,  First  step,  21  -i-  ^  =  37 ; 
Second  step,  $.15  x  27  =  $4.05. 

Find  the  cost  of  the  following: 

2.  124  acres  of  land  at  $144  for  2f  acres ;  for  If  A. 

3.  486  bu.  wheat  at  $11  for  8  bushels ;  at  $4.74  for  3  bushels ; 
at  $.72  for  f  of  a  bushel. 

4.  265  cords  of  wood  at  $21.95  for  5  cords. 

5.  135  yd.  broadcloth  at  $8.97  for  2|  yd.;  at  $12.65  for 
3fyd. 

(217) 


90  BUSINESS    ARITHMETIC, 

6.  What  is  the  cost  of  987  lb.  coal,  at  35  ct.  per  100  lb.  ? 

Solution.— As  the  price  is  per  100  lb.,  we  find  the  number  of  hun- 
dreds is  987  by  moving  the  decimal  point  two  places  to  the  left.  The 
price  multiplied  by  this  result  will  give  the  required  cost.  Hence  $.35  x 
9.87= $3.4545,  the  cost  of  987  lb.  at  35  ct.  per  100  lb. 

Find  the  cost  of  the  following  bill  of  lumber  : 

7.  2345  ft.  at  $1.35  per  100  (0)  feet ;  3628  ft.  at  $.98  per  0. ; 
1843  ft.  at  $1.90  per  0.  ft.  ;  8364  ft.  at  $2.84  per  C. ;  4384  ft. 
at  $27.50  per  1000  (M)  ft.  ;  19364  ft.  at  $45.75  per  M. 

8.  What  is  the  cost  of  84690  lb.  of  coal  at  $6.45  per  ton 
(2000  lb.)  ? 

Observe,  that  pounds  are  changed  to  tons  by  moving  the  decimal  point 
3  places  to  the  left  and  dividing  by  2 

9.  What  is  the  cost  of  96847  lb.  coal  at  $7.84  per  ton  ? 

497.  Prob.  IV. — To  find  the  number  of  units  when 
the  cost  and  price  ol"  one  unit  arc  given. 

1.  How  many  yards  of  cloth  can  be  bought  for  $28  @  $f  ? 
Solution. — Since  1  yard  can  be  bought  for  $|,  as  many  yards  can  be 

bought  for  $28  as  $f  is  contained  times  in  it.  Hence,  $28  -^-  $f  =  49  yd. 

Find  the  price  and  explain  the  following  orally : 

2.  How  many  tons  of  coal  can  be  bought  for  $56  at  $4  a 
ton?    At  $7?    At  $8?    At  $14?     At  $6?    At  $9?    At  $5? 

3.  For  $40  how  many  bushels  of  corn  can  be  bought  at  $f 
perbu.?    At  $4?    At$y\?    At$|J?    At  $.8?    At$|? 

4.  How  many  pounds  of  coffee  can  be  bought  for  $60  at  S^ 
per  pound  ?    At  $f  ?    At  $A  ?    -^^  %^  ?    At  $.33|  ?    At  $.4  ? 

Solve  the  following : 

5.  The  cost  of  a  piece  of  cloth  is  $480,  and  the  price  per 
yard  $1| ;  how  many  yards  does  it  contain  ? 

6.  How  many  bushels  of  wheat  at  $1|  can  be  purchased  for 
$840?     At  $1-1?    At  $4?    At  $14-?    At$l|? 

7.  The  cost  of  digging  a  drain  at  $3f  per  rod  is  $187 ;  what 
is  the  length  of  the  drain  ?  Ans.  51  rd. 

(218)  • 


BUSINESS    PROBLEMS,  91 

8.  A  farmer  paid  $14198  for  his  farm,  at  $65-|  per  acre; 
how  many  acres  does  the  farm  contain  ?  Ans.  217  A. 

9.  A  grocer  purchased  1101. 65  worth  of  butter,  at  35|  cents 
a  pound ;  how  many  pounds  did  he  purchase  ?  A71S.  285  lb. 

10.  How  many  yards  of  cloth  can  be  bought  at  $2.75  a  yard 
for  $1086.25?  A7is.  395. 

11.  A  grain  dealer  purchased  a  quantity  of  wheat  at  11.20 
per  bushel,  and  sold  it  at  an  advance  of  9^^  cents  per  bushel, 
receiving  for  the  whole  $616,896;  how  many  bushels  did  he 
purchase  ? 

498.  Prob.  V. — To  find  the  number  of  units  that  can 
be  purchased  for  a  given  sum  when  the  cost  of  a  multi- 
pie  or  part  of  one  unit  is  given. 

1.  At  19  ct.  for  f  of  a  yard,  how  many  yards  can  be  bought 

for  $8.55  ? 

Solution. — 1.  Since  i]  yd.  cost  19  ct.,  ^  must  cost  |-  of  19  ct.,  or  9|^  ct., 
and  |,  or  1  yard,  must  cost  3  times  9|^  ct.,  or  28|  ct. 

2.  Since  1  yard  cost  28i  ct.,  as  many  yards  can  be  bought  for  $8.55  as 
28  i  ct.  are  contained  times  in  it.  Hence,  $8.55  -;-  $.285  =  30,  the  num- 
ber of  yards  that  can  be  bought  for  $8.55,  at  19  ct.  for  f  yd. 

2.  IIow  many  tons  of  coal  can  be  bought  for  $277.50,  at  $S 
for  f  of  a  ton  ?    At  $8  for  f  of  a  ton  ?  Atis.  37  T. 

3.  How  many  bushels  of  corn  can  be  bought  for  $28,  at 
32  ct.  for  f  of  a  bu.  ?    At  28  ct.  for  |  bu.  ? 

4.  A  town  lot  was  sold  for  $1728,  at  $3  per  8  sq.  ft.  The 
front  of  the  lot  is  48  ft.     What  is  its  depth  ?       Ans,  96  ft. 

5.  A  piece  of  cloth  was  sold  for  $34.50,  at  14  yards  per  $1. 
How  many  yards  did  the  piece  contain?  A71S.  483  yd. 

6.  A  drove  of  cattle  was  sold  for  $3738,  at  $294  for  every 
7  head.    How  many  head  of  cattle  in  the  drove  ?    Aiis.  89. 

7.  A  pile  of  wood  was  bought  for  $275.60,  at  $1.95  for 
3  cord  feet.     How  many  cords  in  the  pile  ?  Ans.  53  cd. 

8.  A  cellar  was  excavated  for  $408.24,  at  $4.41  for  every 
7  cu.  yd.     The  cellar  was  54  ft.  by  36  ft.     How  deep  was  it  ? 

(219) 


2  )  $1.44 
8 

11.52 

2)       72 

4)       36 

9 

Cost  of  8  bu. 
«     "  2  pk. 
«     "  1  pk. 
''     «  2qt. 

92  BUSINESS    ARITHMETIC, 

499.  Prob.  VI. — To  find  the  cost  when  the  quantity 
is  a  compound  number  and  the  price  of  a  unit  of  one 
denomination  is  given. 

1.  What  is  the  cost  of  8  bu.  3  pk.  2  qt  of  wheat,  at  $1.44 

per  bushel  ? 

Solution. — 1.  Since  $1.44  is  the 
price  per  bushel,  $1.44  x  8,  or 
$11.53,  is  the  cost  of  8  bushels. 

2.  Since  2  pk.  =  ^  bu.,  $1.44-r3, 
or  72  cts.,  is  the  cost  of  2  pk.,  and 
the  \  of  72  ct.,  or  36  ct.,  is  the  cost 
of  1  pk. 

3.  Since  there  are  8  qt.  in  1  pk. , 
2  qt.  =  ^  pk.     Hence,  the  cost  of 

$12.69,  Ans.  1  pk.,  36  ct.  -i-  4,  or  9  ct.,  is  the 

cost  of  2  qt. 
4.  The  sum  of  the  cost  of  the  parts  must  equal  the  cost  of  the  whole 
quantity.    Hence,  $12.69  is  the  cost  of  8  bu.  3  pk.  2  qt.,  at  $1.44  per  bu. 

Find  the  cost  of  the  following  orally : 

2.  9  lb.  8  oz.  sugar,  at  12  ct.  per  pouiid  ;  (^  14  ct. ;  ©  20  ct. 

3.  7}  yd.  ribbon  @  16  ct. ;  @  40  ct. ;  @  30  ct. 

4.  15  bu.  3  pk.  6  qt  of  apples,  @,  $1  per  bushel. 

5.  3  lb.  12  oz.  butter,  at  34  ct.  per  pound;  at  40  ct. 
Solve  the  following: 

6.  What  will  5  T.  15  cwt.  50  lb.  sugar  cost,  at  $240  per  ton  ? 

7.  Find  the  cost  of  48  lb.  9  oz.  10  pwt.  of  block  silver,  at 
$12  per  pound.  Ans.  $585.50. 

8.  Find  the  cost  of  excavating  240  cu.  yd.  13^  cu.  ft.  of 
earth,  at  50  cts.  per  cubic  yard. 

9.  How  much  will  a  man  receive  for  2  yr.  9  mo.  25  da. 
service,  at  $1800  per  year?  Ans.  $5075. 

10.  Sold  48  T.  15  cwt.  75  lb.  of  hay  at  $15  per  ton,  and 
32  bu.  3  pk.  6  qt.  timothy  seed  at  $3.50  per  bushel.  How 
much  did  I  receive  for  the  whole  ? 

11.  How  much  will  it  cost  to  grade  8  mi.  230  rd.  of  a  road, 
at  $4640  per  mile  ?  Ans.  $40455. 

(220) 


BUSINESS    PROBLEMS,  93 

500.  Prob.  VII. — To  find  what  part  one  number  is  of 
another. 

1.  What  part  of  12  is  4? 

Solution. — 1  is  ^^  of  13  and  4  beiifg  4  times  1,  is  4  times  ^V  of  13, 
which,  is  ^^  =  ^  ;  hence  4  is  |  of  13. 

Observe,  that  to  ascertain  what  part  one  number  is  of  another,  we  may 
at  once  write  the  former  as  the  numerator  and  the  latter  as  the  denom- 
inator of  a  fraction,  and  reduce  the  fraction  to  its  lowest  terms  (250). 

2.  What  part  is  15  of  18  ?    Of  25  ?    Of  24  ?     Of  45  ? 

3.  What  part  is  36  of  48?    Of  38?    Of  42?    Of  72? 

4.  I  is  what  part  of  ^  ? 

Solution. — 1.  Only  units  of  the  same  integral  and  fractional  denom- 
ination can  be  compared  (155) ;  hence  we  reduce  |  and  f  to  |f  and  ^f, 
and  place  the  numerator  14  over  the  numerator  18,  giving  il  =  I ;  hence 
f  is  i  of  f . 

3.  We  may  express  the  relation  of  the  fractions  in  the  form  of  a 
complex  fraction,  and  reduce  the  result  to  a  simple  fraction  (300).  Thus 

f  =  tI  =  I-     Hence  f  is  |  of  f . 

5.  f  is  what  part  of  11  ?    ^  is  what  part  of  2J? 

6.  5f  inches  is  what  part  of  2^  yards  ?    (See  387.) 

7.  29^  rods  is  what  part  of  1  mile  ? 

8.  7^  is  how  many  times  f  ? 

9.  llf  is  how  many  times  2|  ? 

10.  What  part  of  a  year  is  24  weeks  ?    8  weeks  10  days  ? 

11.  A  man's  yearly  wages  is  $950,  and  his  whole  yearly  ex- 
penses 8590.80.  What  part  of  his  wages  does  he  save  each 
year  ? 

12.  Out  of  $750  I  paid  $240.  What  part  of  my  money 
have  I  still  left  ?  Ans.  ^,  or  .68. 

13.  A  man  owning  a  farm  of  240|  acres,  sold  117-5-  acres. 
What  part  of  his  whole  farm  has  he  still  left  ? 

14.  4^  is  what  part  of  12^  ?     S%  is  what  part  of  14^  ? 

15.  3|^  is  what  part  of  9%  ?     7i%  is  what  part  of  8|^  ? 

(221) 


94  BUSINUSS    ARITHMETIC, 

16.  Illustrate  in  full  the  process  in  the  14th  and  15th 
examples. 

501.  Peob.  VIIL— To  find  a  given  fractional  part  of  a 
given  number. 

1.  Find  -I  of  238. 

Solution.— We  find  j  of  238  by  dividing  it  by  7  ;  hence  238^7  =  34, 
the  1-  of  238.     But  f  is  3  times  j  ;  hence  34  x  3=102,  the  f  of  238. 

2.  Find  J  of  48  ;  of  96 ;  of  376  ;  of  1035. 

3.  Find  ^-^  of  340 ;  ^f  ^  of  972 ;  ^  of  560. 

4.  Find  ^  of  $75 ;  ^  of  1824.60  ;  ^^  of  13.25. 

Observe,  that  f  of  $75  means  such  a  number  of  dollars  as  will  contain 
$4  for  every  $5  in  $75  ;  hence,  to  find  the  f  of  $75,  we  divide  by  5  and 
multiply  the  quotient  by  4. 

5.  Find  11%  of  328. 

Solution. — 1.  7%  means  y^tt*  ^®  ^^^  toit  ^7  moving  the  decimal 
point  two  places  to  the  left  (4:70).  Hence  7^/o  or  y^^  of  328  is  equal 
to  3.28  X  7  =  22.96. 

2.  We  usually  multiply  by  the  rate  first,  then  point  off  two  decimal 
places  in  the  product,  which  divides  it  by  100. 

6.  What  is  8%  of  1736  ?    4^  of  395  lb.  butter  ? 

21  2^ 

7.  How  much  is  -^  of  157  acres  ?    -^f  of  84  bu.  wheat  ? 

o  7 

Find  Find 

8.  7%  of  28  yd.  11.  ^%  of  284  mi. 

9.  5%  of  300  men.  12.  12^^  of  732. 
10.  9%  of  278  lb.  13.  f  ^  of  $860. 

14.  Find  the  amount  of  1832  +  i%  of  itself. 

15.  Find  the  amount  of  $325  +  7%  of  itself 

16.  A  firkin  of  butter  contained  72|  lb. ;  f  of  it  was  sold: 
how  many  pounds  are  there  left  ? 

17.  A  piece  of  cloth  contained  142  yd. ;  15^  was  sold  :  how 
many  yards  yet  remained  unsold  ? 

(222) 


BUSIJVL'SS    PROBLEMS,  95 

18.  James  Smith's  farm  contained  284  acres,  and  H.  A. 
Watkins'  ^%  less.     How  many  acres  in  H.  A.  Watkins'  farm  ? 

19.  A  merchant  bought  276  yards  cloth  at  $3.40  per  yard. 
He  sold  it  at  %b%  profit.  How  much  did  he  reahze,  and  what 
was  his  selling  price  ? 

20.  If  tea  cost  96  ct.  per  pound  and  is  sold  at  a  loss  of  l^%, 
what  is  the  selling  price  ? 

503.  Peob.  IX. — To  find  a  number  when  a  fractional 
part  is  given. 

1.  Find  the  number  of  which  84  is  }. 

Solution.— Since  84  is  |  of  the  number,  |  of  84  must  be  ^  ;  hence 
84  -5-  7  =  13  is  the  ^  of  the  required  number.  But  9  times  f^  is  equal  to 
the  whole  ;  hence,  12  x  9  =:  108,  the  required  number. 

2.  $36  is  f  of  how  many  dollars  ?  $49  is  |  of  how  many 
dollars  ? 

3.  Find  the  number  of  yards  of  cloth  of  which  135  yd.  is  -f^, 

4.  James  has  $756, which  is  ^  of  George's  money ;  how  many 
dollars  has  George  ?  Ans.  $1323. 

5.  The  profits  of  a  grocery  for  one  year  are  13537,  which  is 
■^^  of  the  capital  invested.    How  much  is  the  capital  ? 

6.  Find  the  number  of  dollars  of  which  $296  are  %%,  or  .08. 

First  Solution.— Since  $296  are^f^  of  the  number,^ of  $296 or  $37, 
are  yi^ ;  hence  \^%,  or  the  whole,  is  100  times  $37,  or  37  x  100  -  $3700. 

Second  Solution. — Since  $296  are  yf  ^  of  the  number,  \  of  $296  is 
yJ^,  and  \  of  100  times  $296  is  \^%,  or  the  required  number.  Hence, 
$296  X  100  =  $29600,  and  |  of  $29600  =  $3700,  the  required  number. 

From  these  solutions  we  obtain  the  following  rule  for  finding  a 
number  when  a  decimal  part  of  it  is  given  : 

503.  Rule. — Move  the  deeimal  point  as  many  places 
to  the  7'ight  as  there  are  places  in  the  given  decimal, 
annexing  ciphers  if  necessary,  and  divide  the  result  hy 
the  number  expressed  hy  the  significant  figures  in  the 
given  decimal. 

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96  BUSINESS    ARITH3IETIC, 

Find  what  number  Find  what  number  Find  what  number 

7.  16  is  8%  of.  13.  f  is  H  of.  19.  ^  is  9^  of. 

8.  24  is  6^  of.  14.  $|  is  7^  of.         20.  12.16  is  6^  of. 

9.  84  is  %  of.  15.  I  pk.  is  8^  of.     21.  7-J  is  9^  of. 

10.  172  are  9^  of.  16.  .7  ft.  is  b%  ol      22.  %^  is  5^  of. 

11.  120  yd.  are  b%  of.      17.  .09  is  4^  of.        23.  f  yd.  is  8^  of. 

12.  56  bu.  are  8^  of.       18.  .48  is  12^  of.       24.  .96  is  12^  of. 

25.  A  man's  profits  for  one  year  amount  to  12840,  which  is 
8^  of  the  amount  he  has  invested  in  business.  What  is  his 
investment  ? 

26.  A  merchant  sells  a  piece  of  cloth  at  a  profit  of  30  ct.  a 
yard,  which  is  20^  of  what  it  cost  him.  What  was  the  buying 
price  per  yard  ? 

27.  A  grocer  purchased  186  lb.  butter  on  Saturday,  which  is 
%%  of  the  entire  quantity  purchased  during  the  week.  What 
was  the  week's  purchase  ? 

28.  A  mechanic  pays  112  a  month  for  house  rent,  which 
is  16%  of  his  wages.    What  does  he  receive  per  month  ? 

29.  12^  of  f  is  9^  of  what  number  ? 

30.  How  many  acres  in  a  farm  14^  of  which  contains 
42  acres  ? 

31.  An  attorney  receives  11.75  for  collecting  a  bill,  which 
ifi  2^  per  cent,  of  the  bill.     What  is  the  amount  of  the  bill  ? 

32.  A  man  having  failed  in  business  is  allowed  to  cancel 
his  debts  by  paying  20^.  What  does  he  owe  a  man  Avho 
receives  $270?  Ans.  $1350. 

33.  A  man  sold  his  house  for  $1000,  which  was  12^  of  the 
sum  he  received  for  his  farm.  What  was  the  price  of  the 
farm  ?  Ans.  $8333.33^. 

34.  If  in  a  certain  town  $3093.75  was  raised  from  a  ^%  tax, 
what  was  the  value  of  property  in  the  town  ? 

35.  S.  T.  Esty  has  25%  of  his  property  invested  in  a  house, 
10%  in  a  farm,  5%  in  a  barn,  and  the  rest  in  a  grove  worth 
$4800.    What  is  the  amount  of  his  property? 

(224) 


BUSINESS    PROBLEMS.  97 

504.  Prob.  X. — To  express  the  part  one  number  is 
of  another  in  any  given  fractional  unit. 

1.  How  many  fifths  oi  3  is  8. 

Solution.  — Since  |  is  ^  of  3,  there  must  be  as  msLny  fifths  of  3  in  8 

as  I  is  contained  times  in  it.     8  -i-  f  =  13|.    Hence,  8  is  -~  of  three. 

5 

Solve  the  following  orally : 

2.  Kow  msLnj  fourths  of  9  is  H  ?    Is  5  ?    Is  12?    Is  20  ?  ' 

3.  How  many  hundredths  of  36  is  9  ?   Is  4  ?  Is  18  ?  Is  12  ? 

4.  $12  are  how  many  tenths  of  $5  ?    Of  18  ?    Of  $15  ? 

5.  42  yards  are  how  many  sixths  of  2  yd.  ?  Of  7  yd.  ?  Of 
3  yd.? 

6.  What  per  cent  of  $11  are  $3,  or  $3  are  how  many  hun- 
dredths of  $11  ? 

FiBST  Solution.— Since  y^V  is  yj^  of  $11,  there  must  be  as  many 
hundredths  of  $11  in  $3  as  yV?y  is  contained  times  in  $3.    $3  -:-  iVt7  = 

3  X  -V^  =  -\o^  =  27tV    Hence,  $3  are  ^-^1,  or  27j\%  of  $11. 

Second  Solution.— Since  (500)  $3  are  j\  of  $11,  we  have  only  to 
reduce  ^  to  hundredths  to  find  what  per  cent  $3  are  of  $11.     y\  =  yWir 

=  ^  =  27^j % .     Hence,  $3  are  27^%  of  $11. 

From  these  solutions  we  obtain  the  following  rule  for  find- 
ing what  per  cent  or  what  decimal  part  one  number  is  of 
another : 

505.  Rule. — Express  the  former  number  as  a  frae- 
tion  of  the  latter  (500),  and  reduce  this  fraction  to 
hundredths  or  to  the  required  decinial  (338). 

Find  what  per  cent 

7.  16  is  of  64.  13.  284  acres  are  of  1  sq.  mi. 

8.  12  is  of  72.  14.  2  bu.  3  pk.  are  of  28  bu. 

9.  $36  are  of  $180.  15.  $f  is  of  $| ;  of  $5 ;  of  $2|. 

10.  $46  are  of  $414.  16.  3  lb.  13  oz.  are  of  9  lb. 

11.  7  feet  are  of  8  yards.      17.  4  of  a  cu.  ft.  is  of  1  cu.  yd. 

12.  13  oz.  are  of  5  lb.  18.  48  min.  are  of  3  hr. 

(225) 


98  BUSI^YESS     ARITHMETIC. 

19.  f  of  a  sq.  yd.  is  of  -|  of  a  sq.  yd. 

20.  3  bu.  2  pk.  are  of  8  bu.  3  pk.  5  qt. 

21.  A  man  paid  $24  for  the  use  of  $300  for  one  year.  What 
rate  per  cent  did  he  pay  ? 

22.  A  merchant  invested  $3485  in  goods  which  he  had  to 
sell  for  $2973.    What  per  cent  of  his  investment  did  he  lose  ? 

23.  A  druggist  paid  84  ct.  an  ounce  for  a  certain  medicine, 
and  sold  it  at  $1.36  an  ounce.  What  per  cent,  profit  did  he 
make? 

SoLUTiON.-$1.36  -  $.84  =.  $.52  ;  ff  =  ff^^  =  ^  =  ^^i^fo- 

24.  A  farmer  owning  386  acres  sold  148  acres.  What  per 
cent  of  his  original  farm  does  he  still  own  ? 

25.  When  a  yard  of  silk  is  bought  for  $1.20  and  sold  for 
$1.60,  what  per  cent  is  the  profit  of  the  buying  price  ? 

26.  A  man  owed  me  $350,  but  fearing  he  would  not  pay  it 
I  agreed  to  take  $306.25 ;  what  per  cent,  did  I  allow  him  ? 

27.  Hawkins  deposited  $2500  in  a  bank,  and  again  deposited 
enough  to  make  the  whole  amount  to  $2750.  What  per  cent 
of  the  first  deposit  was  the  last  ?  Ans.  10. 

28.  Gave  away  77^  bushels  of  potatoes,  and  my  whole  crop 
was  500  bushels  ;  what  %  of  the  crop  did  I  give  away  ? 

29.  A  man  pays  $215.34  per  acre  for  4J  acres  of  land,  and 
lets  it  a  year  for  $33.916 ;  what  %  of  the  cost  is  the  rent  ? 

506.  Prob.  XL — To  find  a  number  which  is  a  given 
fraction  of  itself  greater  or  less  than  a  given  number. 

1.  'Find  a  number  which  is  f  of  itself  less  than  28. 

Solution. — 1.  Since  the  required  number  is  f  of  itself,  and  is  f  of 
itself  less  than  28,  hence  28  is  f +  f  or  |  of  it. 

2.  Since  28  is  |  of  tlie  number,  \  of  28,  or  4,  is  J.  Hence  f ,  or  the 
whole  of  the  required  number,  is  5  times  4  or  20. 

Solve  the  following  orally : 

2.  What  number  is  f  of  itself  less  than  15  ?  Less  than  40  ? 
Less  than  75  ?    Less  than  ^6  ?    Less  than  32  ? 

(226) 


B  [/SIJVJSSS     PROBLEMS.  99 

3.  What  number  increased  f  of  itself  is  eqnal  100  ?  Is  equal 
80  ?     Is  equal  120  ?     Is  equal  12  ?     Is  equal  7  ? 

4.  Find  a  number  which  diminished  by  f  of  itself  is  equal 
50.     Is  equal  70.     Is  equal  15.     Is  equal  5. 

Solve  and  explain  the  following: 

5.  What  number  increased  by  7%  or  yJo-  of  itself  is  equal 
G42? 

Solution.— 1.  Since  a  number  increased  by  7%  or  jl^  of  itself  is 
tSs  +  t^tt  =  lU  o^  itself,  642  is  \^}  or  107%  of  the  required  number. 

2.  Since  G42  is  [g J  of  the  required  number,  for  every  107  in  643  there 
must  be  100  in  the  required  number.  Hence,  642  -r- 107  =  6,  and  6  x  100 
=  600,  the  required  number. 

Observe,  that  642  -i-  1.07  is  the  same  as  dividing  by  107  and  multiply- 
ing by  100  (360) ;  hence  the  following  rule,  when  a  number  has  been 
increased  or  diminished  by  a  given  per  cent  or  any  decimal  of  itself : 

50*7.  Rule. — Dimde  the  given  number,  according  as 
it  is  more  or  less  than  the  required  iiumher,  dy  1  in- 
creased or  diminished  by  the  given  decimal. 

6.  What  number  increased  by  15;^  of  itself  is  equal  248.40? 

7.  A  certain  number  increased  by  %0%  of  itself  is  331.2 ; 
what  is  that  number  ?  Ans.  184. 

8.  By  running  15^  faster  than  usual,  a  locomotive  runs  044 
miles  a  day ;  what  was  the  usual  distance  per  day  ? 

9.  What  number  diminished  by  25^  of  itself  is  654  ? 

10.  A  regiment  after  losing  %%  of  its  number  contained  736 
men  ;  what  was  its  original  number  ?  Ans.  800. 

11.  A  man  who  has  had  his  salary  increased  5^  now  receives 
$1050  a  year ;  what  was  his  former  salary  ?        Ans.  81000. 

12.  A  merchant  sells  a  coat  for  $8,  thereby  gaining  25;^  ; 
what  did  the  coat  cost  him  ?  A7is.  $6.40. 

13.  A  clergyman  lays  up  12|^  of  his  salary,  which  leaves 
him  $1750  to  spend;  what  is  his  salary?  Ans.  $2000. 

14.  J.  Fayette  sold  his  farm  for  83960,  which  was  10;^  less 
than  he  gave  for  it,  and  he  gave  10;^  more  than  it  was  worth ; 
what  was  its  actual  value  ?  Ans,  $40.00. 

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100  BUSINESS     ARITHMETIC, 


APPLICATIONS. 

508.  Profit  and  Loss,  Commission,  Insurance,  Stocks, 
Taxes,  and  Duties,  are  applications  of  Business  Problems 
VIII,  IX,  X,  XI.  The  rate  in  these  subjects  is  usually  a 
jper  cent.  Hence,  for  convenience  in  expressing  rules,  we 
denote  the  quantities  by  letters  as  follows: 

1.  H  represents  the  Base,  or  number  on  which  the  percentage  is 
reckoned. 

2.  It  represents  the  Rate  per  cent  expressed  decimally. 

3.  JP  represents  the  Percentage,  or  the  part  of  the  Base  which  is 
denoted  by  the  Rate. 

4.  A  represents  the  Amount,  or  sum  of  the  Base  and  Percentage. 

5.  D  represents  the  Difference,  or  Base  less  the  Percentage. 

Formulce,  or  Mules  for  Percentage, 

509.  Prob.  VIU.    P  =  B  X  B.  Read,  \  ^"'  V^^'^^ye  ismf  to  th, 

{     base  multiplied  hy  the  rate. 

--i/v_,T^        x>      J*  -D    A    \  '^^^^  ^^*^  ^  equal  to  the  per- 

510.  Prob.  IX.       B  =  -^.         Read, -^  ^-  -^  ^  ,>    w       . 

R  {     centage  divided  hy  the  rate. 

K-t-i-r.^^  ^-P  T>j(  TJie  rate  is  equal  to  the  per- 

511.  Prob.  X.        -R  =  -B-  Read,-^  ^-  v,  ^  t.    ^x    , 

R  (      centage  divided  by  the  base. 

i  Tt—      ^  T?     d    i  ^^^(^^^^^^Q^^^^^i^^^^^Of^f^^ 

rt-io     Tit-xTT")       ~  1  +  It'  {      divided  by  1  plus  the  rate. 

OVa.      Prob.  XI.      )  _^  /    mr     7.  .  7.     .7      3.^. 

i  _.         X)       T>     J    i  Thebaseiscqualtothedtffnce 
{  1  —  R  {      divided  by  1  minus  tTie  rate. 

513.  Refer  to  the  problems  on  pages  222  to  226  inclusive, 
and  answer  the  following  questions  regarding  these  formulae : 

1.  What  is  meant  hy  BxE,  and  why  is  P=B  xR9  Illus- 
trate your  answer  by  an  example,  giving  a  reason  for  each 
step. 

2.  Why  is  P  -^  E  equal  B  ?  Give  reasons  in  full  for  your 
answer. 

3.  If  i2  is  135^,  which  is  the  greater,  P  or  By  and  why  ? 

(228) 


Bisrn  ^? 


PRO  Fir    AND     LOSS.  101 

4.  If  R  is  248^,  how  would  you  express  R  without  the 

5.  Why  is  R  equal  to  P  -r-  ^,  and  how  must  the  quotient 
of  P  -7-  i5  be  expressed  to  represent  R  correctly  ? 

G.  What  is  meant  by  ^  ?  How  many  times  R  in  P  (502)  ? 
How  many  times  1  in  P  ?  How  many  times  1  +  R  must 
there  be  in  A  and  why  ? 

7.  How  many  times  R  in  P  (503)?  D  is  equal  to  B 
minus  liow  many  times  R  (501)  ? 

8.  Why  is  B  equal  to  D  --  {I  —  R)?  Give  reasons  in  full 
for  your  answer. 

PEOFIT  AND  LOSS. 

514.  The  quantities  considered  in  Profit  and  Loss  corres- 
pond with  those  in  Percentage  thus : 

1.  The  Cost,  or  Capital  invested,  is  the  Base, 

2.  The  I^er  cent  of  Profit  or  Loss  is  the  Hate. 

3.  The  JProfit  or  Loss  is  the  Percentage. 

4.  The  Selling  Price  when  equal  the  Cost  plus  the 
Frojit  is  the  Amount,  when  equal  the  Cost  minus  the  Loss  is 
the  Uifference. 

EXAMPLKS     FOR     PRACTICES. 

515.  1.  A  firkin  of  butter  was  bought  for  119  and  sold  at 
a  profit  of  1Q%.  What  was  the  gain  ? 

Formula  P  =  B  x  R.    Read,  Profit  or  Loss  —  Cost  x  Bate  %. 
Find  the  profit  on  the  sale 

2.  Of  320  yd.  cloth  bought  @  $1.50,  sold  at  a  gain  of  11%. 

3.  Of  84  cd.  wood  bought  @  $4.43^,  sold  at  a  gain  of  20^. 

4.  Of  873  bu.  wheat  bought  @  $1.25,  sold  at -a  gain  of  14J^. 

Find  the  loss  on  the  sale 

5.  Of  180  T.  coal  bought  @  17.85,  sold  at  a  loss  of  %\%, 

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102  BUSINESS     ARITHMETIC. 

6.  Of  124  A.  land  bought  @  $84.50,  sold  at  a  loss  of  21|^. 

7.  If  a  farm  was  bought  for  $4860  and  sold  for  $729  more 
than  the  cost,  what  was  the  gain  per  cent 

Formula  R  =  P  -i-  B.    Read,  Rate  %  Gain  =  Profit  -i-  Cost, 

8.  A  piece  of  cloth  is  bought  at  $2.85  per  yard  and  sold  at 
$2.10  per  yard.    What  is  the  loss  per  cent  ?  Ans. 

9.  If  f  of  a  cord  of  wood  is  sold  for  f  of  the  cost  of  1  cord, 
what  is  the  gain  per  cent  ?  Ans. 

10.  Find  the  selling  price  of  a  house  bought  at  $5385.90, 
and  sold  at  a  gain  of  18^. 

Formula  A=B  x  (1  +  R).  Read,  Selling  Price =Ci?si  x  (1  +  Rate  %  Oain). 

11.  Corn  that  cost  65  ct.  a  bushel  was  sold  at  20^  gain. 
What  was  the  selling  price  ?  Ans.  78  ct.  a  bu. 

12.  A  grocer  bought  43  bu.  clover  seed  @  14.50,  and  sold  it 
in  small  quantities  at  a  gain  of  40^.  What  was  the  selling 
price  per  bu.  and  total  gain  ? 

13.  Bought  184  barrels  of  flour  for  $1650,  and  sold  the  whole 
at  a  loss  of  S%.    What  was  the  selling  price  per  batrel  ? 

Formula  D = B  (1  -  R).     Read,  Selling  Price  =  Cost  x  (1  -  Rate  %  Loss), 

14.  Flour  was  bought  at  $8.40  a  barrel,  and  sold  so  as  to 
lose  15^.    What  was  the  selling  price? 

15.  0.  Baldwin  bought  coal  at  $6.25  per  ton,  and  sold  it  at 
a  loss  of  18^.    What  was  the  selling  price  ? 

16.  Sold  a  house  at  a  loss  of  $879,  which  was  15^  of  the 
cost.     What  was  the  cost  ? 

Formula  B  =  P  -j-  R.    Read,  Cost  =  Profit  or  Loss  -i-  Rate  %. 

17.  A  grain  merchant  sold  284  barrels  of  flour  at  a  loss  of 
$674.50,  which  was  25^  of  the  cost.  Wliat  was  the  buying 
and  selling  price  per  barrel  ? 

18.  A  drover  wished  to  realize  on  the  sale  of  a  flock  of 
236  sheep  $531*  which  is  30%  of  the  cost.  At  what  price  per 
head  must  he  sell  the  flock? 

19.  Two  men  engaged  in  business,  each  having  $4380.    A 

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COMMISSION.  103 

gained  33|^  and  B  75^.    How  much  was  B's  gain  more 
than  A's? 

20.  If  I  buy  72  head  of  cattle  at  136  a  head,  and  sell  33^^ 
of  them  at  a  gain  of  18^,  and  the  remainder  at  a  gain  of  24^, 
what  is  my  gain  ? 

21.  A  grocer  sells  coffee  that  costs  13|^  cents  per  pound, 
for  lOf  cents  a  pound.     What  is  the  loss  per  cent  ? 

22.  Fisk  and  Gould  sold  stock  for  $3300  at  a  profit  of  ^^%, 
What  was  the  cost  of  it  ? 

23.  A  man  bought  24  acres  of  land  at  175  an  acre,  and  sold 
it  at  a  profit  of  8  J^.     What  was  his  total  gain  ? 

24.  A  merchant  sold  cloth  for  $3.84  a  yard,  and  thus  made 
20^.     What  was  the  cost  price  ? 

25.  Bought  wood  at  $3.25  a  cord,  and  sold  it  at  an  average 
gain  of  30^.     What  did  it  bring  per  cord  ? 

26.  If  land  when  sold  at  a  loss  of  l^%  brings  111.20  per 
acre,  what  would  be  the  gain  per  cent  if  sold  for  $15.36  ? 

27.  Bought  a  barrel  of  syrup  for  $20  ;  what  must  I  charge 
a  gallon  in  order  to  gain  20^  on  the  whole? 


COMMISSION. 

516.  A  Commission  Merchant  or  Agent  is  a  per- 
son who  transacts  business  for  another  for  a  percentage. 

517.  A  Broker  is  a  person  who  buys  or  sells  stocks,  bills 
of  exchange,  etc.,  for  a  percentage. 

518.  Commission  is  the  amount  paid  a  commission 
merchant  or  agent  for  the  transaction  of  business. 

519.  lirokeraffe  is  the  amount  paid  a  broker  for  the 
transaction  of  business. 

530.  The  Net  I^roceeds  of  any  transaction  is  the  sum 
of  money  that  is  left  after  all  expenses  of  commission,  etc., 
are  paid. 

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104  BUSINUSS    ARITHMETIC, 

531.  The  quantities  considered  in  commission  correspond 
with  those  in  percentage  thus  : 

1.  The  amount  of  money  invested  or  collected  is  the  Base. 

2.  The  per  cent  allowed  for  services  is  the  Hate. 

3.  The  Commission  or  Brokerage  is  the  I*erc€ntaf/e. 

4.  The  sum  invested  or  collected,  plus  the  commission,  is  the 
Aiiioimt,  minus  the  commission  is  the  Difference. 

EXAMPIiES     FOR     PRACTICE. 

522.  Let  the  pupil  write  out  the  formulae  for  each  kind 
of  examples  in  commission  in  the  same  manner  as  they  are 
given  in  Profit  and  Loss. 

What  is  the  commission  or  brokerage  on  the  following : 

1.  The  sale  of  85  cords  of  wood  @  $4.75,  commission  3^%  ? 

2.  The  sale  of  484  yds.  cloth  @  $2.15,  commission  1}^  ? 

3.  The  sale  of  176  shares  stocks  at  187.50  a  share,  broker- 
age i%  ? 

4.  The  collection  of  13462.84,  commission  2^%  ? 

What  is  the  rate  of  commission  on  the  following : 

5.  Selling  a  farm  for  $4800,  commission  1120  ? 

6.  Collecting  a  debt  of  $7500,  commission  $350  ? 

7.  Selling  wheat  worth  $1.80  a  bu.,  commission  4  ct.  a 
bushel  ? 

What  is  the  amount  of  the  sale  in  the  following : 

8.  The  commission  is  $360,  rate  of  commission  2 J^  ? 

9.  The  brokerage  is  $754.85,  rate  of  brokerage  1^%  ? 

10.  The  commission  is  $26.86,  rate  of  commission  If  ^  ? 

Find  the  amount  of  the  sales  in  the  following: 

Observe,  that  the  commission  is  pn  the  amount  of  the  sales.  Hence  the 
formula  for  finding  the  amount  of  the  sales  when  the  net  proceeds  are 
given  is  (506) 

Amount  of  sales  =  Wet  proceeds  ^  (1  —  Bate  %). 

11.  Net  proceeds,  $8360  ;  rate  of  commission,  3^%. 

(232) 


INSURANCE,  105 

12.  Net  proceeds,  $3G40 ;  rate  of  commission,  |^. 

13.  Net  proceeds,  $1850;  rate  of  commission,  f^. 

Find  the  amount  to  be  invested  in  the  following : 

Observe,  that  when  an  agent  is  to  deduct  his  commission  from  the 
amount  of  money  in  his  hand  the  formula  is  (500) 

Sum  invested  =  Amount  in  hand  -r-  (1  +  Rate  % ). 

14.  Amount  in  hand,  $3401.01 ;  rate  of  commission,  3J^. 

15.  Eemittance  was  $393.17  ;  rate  of  commission,  2f^. 

16.  Amount  in  hand,  $606.43;  rate  of  commission,  IJ^. 

17.  A  lawyer  collects  bills  amounting  to  $492;  what  is  his 
commission  at  5^  ?  ^7i5.  $24.60. 

18.  An  agent  sold  824  barrels  of  beef,  averaging  202 J  lb. 
each  at  9  cents  a  pound ;  what  was  his  commission  at  2 J^  ? 

19.  A  merchant  has  sent  me  $582.40  to  invest  in  apples,  at 
$5  a  barrel ;  how  many  can  I  buy,  commission  being  4^  ? 

20.  I  have  remitted  $1120  to  my  correspondent  in  Lynn  to 
invest  in  shares,  after  deducting  his  commission  of  1^% ;  what 
is  his  commission  ? 

21.  An  auctioneer  sold  goods  at  auction  for  $13825,  and 
others  at  a  private  sale  for  $12050;  what  was  his  commission 
at  J^?  Ans.  $129.3750. 

22.  A  man  sends  $6897.12  to  his  agent  in  New  Orleans, 
requesting  him  to  invest  in  cotton  after  deducting  his  com- 
mission of  2^;  what  was  the  amount  invested? 


INSUEANCE. 

533.  Insurance  is  a  contract  which  binds  one  party  to 
indemnify  another  against  possible  loss  or  damage.  It  is  of 
two  kinds :  insurance  on  property  and  insurance  on  life, 

524.  The  Policy  is  the  written  contract  made  between 
the  parties. 

535.  The  I^remium  is  the  percentage  paid  for  insurance. 

(233) 


106  BUSIJV^ESS     ARITHMETIC. 

526.  The  quantities  considered  in  insurance  correspond 
with  those  in  percentage ;  thus, 

1.  The  amount  insured  is  the  Base, 

2.  The  jicr  cent  of  premium  is  the  Hate, 

3.  The  premium  is  the  Percentage, 

EXAMPLES     FOR      PRACTICE. 

537.  Let  the  pupil  write  out  the  formulae  as  in  Profit  and 
Loss. 

1.  What  is  the  premium  on  a  policy  for  $3500,  at  3^  ? 

2.  My  house  is  insured  for  $7250 ;  what  is  the  yearly  pre- 
mium, at  2f  ^  ? 

3.  Justus  Weston's  house  is  insured  for  $3250  at  3|-  per 
cent,  his  furniture  for  $945  at  1}  per  cent,  and  his  bam  for 
$1220  at  1^  per  cent ;  what  is  the  amount  of  premium  on  the 
whole  property  ? 

4.  A  factory  is  insured  for  $27430,  and  the  premium  is 
$685.75 ;  what  is  the  rate  of  insurance  ? 

5.  The  Pacific  Mills  of  Lawrence,  worth  $28000,  being 
insured  for  f  their  value,  v/ere  destroyed  by  fire;  at  2f  per 
cent,  what  is  the  actual  loss  of  the  insurance  company  ? 

6.  The  premium  on  a  house,  afc  f  per  cent,  is  $40  ;  what  is 
the  sum  insured  ? 

7.  It  costs  me  $72  annually  to  keep  my  house  insured 
for  $18000 ;  what  is  the  rate  ? 

8.  What  must  be  paid  to  insure  from  Boston  to  New  Or- 
leans a  ship  valued  at  $37600,  at  |  of  1^  ? 

9.  A  cargo  of  800  bundles  of  hay,  worth  $4.80  a  bundle,  is 
insured  at  1^%  on  J  of  its  full  value.  If  the  cargo  be  destroyed, 
how  much  will  the  owner  lose  ? 

10.  My  dwelling-house  is  insured  for  $4800  at  Y/o  ;  ^ij  f i^r- 
niture,  library,  etc.,  for  $2500  at  ^% ;  my  horses,  cattle,  etc., 
for  $3900  at  Y/o  >  ^^^^  ^  carriage  manufactory,  including 
machinery,  for  $4700  at  1J%'.    What  is  my  annual  premium? 

(234) 


T0CK8,  107 


STOCKS. 


528.  A  Corporation  is  a  body  of  individuals  or  com- 
pany authorized  by  law  to  transact  business  as  one  person. 

529.  The  Capital  Stock  is  the  money  contributed  and 
employed  by  the  company  or  corporation  to  carry  on  its 
business. 

Tlie  term  stock  is  also  used  to  denote  Government  and  State  bonds,  etc. 

530.  A  Share  is  one  of  the  equal  parts  into  which  the 
mpital  stock  is  divided. 

531.  A  Certificate  of  Stocky  or  Scripy  is  a  paper 
issued  by  a  corporation,  securing  to  the  holder  a  given  num- 
ber of  shares  of  the  capital  stock. 

532.  The  Par  Value  of  stock  is  the  sum  for  which  the 
Bcrip  or  certificate  is  issued. 

533.  The  Market  Value  of  stock  is  the  price  per  share 
for  which  it  can  be  sold. 

534.  The  Premium^  Discount^  and  Brokerage 

are  always  computed  on  t\\Q  par  value  of  the  stock. 

535.  The  Wet  Earnings  are  the  moneys  left  after 
deducting  all  expenses,  losses,  and  interest  upon  borrowed 
capital. 

536.  A  Bond  is  a  written  instrument,  securing  the  pay- 
ment of  a  sum  of  money  at  or  before  a  specified  time. 

537.  A  Coupon  is  a  certificate  of  interest  attached  to  a 
bond,  which  is  cut  off  and  delivered  to  the  payor  when  the 
interest  is  discharged. 

538.  U".  S.  Bonds  may  be  regarded  as  of  two  classes : 
those  payable  at  a  fixed  date,  and  those  payable  at  any  time 
between  two  fixed  dates,  at  the  option  of  the  government. 

(235) 


108  BUSINESS    ARITHMETIC, 

539.  In  commercial  language,  the  two  classes  of  TJ.  S. 
bonds  are  distinguished  from  each  other  thus : 

(1.)   U,  S,  O's,  bonds  payable  at  a  fixed  time. 

(3.)  U.  S.  6's  5-20,  bonds  payable,  at  the  option  of  the  Govern- 
ment, at  any  time  from  5  to  20  years  from  their  date. 

EXAMPLES     FOR     PRACTICE. 

540.  Let  the  pupil  write  out  the  formula  for  each  class  of 
examples,  as  shown  in  Profit  and  Loss : 

1.  Find  the  cost  of  120  shares  N.  Y.  Central  stock,  the 
market  value  of  which  is  108,  brokerage  \%. 

Solution. -Since  1  share  cost  108%  +1%,  or  108|%  of  $100  =  108|, 
the  cost  of  120  shares  will  be  $108|  x  120  =  $13020. 

2.  What  is  the  market  value  of  86  shares  in  the  Salem  and 
Lowell  Eailroad,  at  d^%  premium,  brokerage  1%  ? 

3.  Find  the  cost  of  95  shares  bank  stock,  at  Q%  premium, 
brokerage  |^. 

4.  How  many  shares  of  Erie  Eailroad  stock  at  8^  discount 
can  be  bought  for  17030,  brokerage  ^%  ? 

Solution.— Since  1  share  cost  100%— 8% +^%,  or  92i%  of  $100  = 
$92.50,  as  many  shares  can  be  bought  as  $92.50  are  contained  times  in 
$7020,  which  is  76. 

How  many  shares  of  stock  can  be  bought 

5.  For  $10092,  at  a  premium  of  b%,  brokerage  \%? 

6.  For  $13428,  at  a  discount  of  7^,  brokerage  \%  ? 

7.  For  $16830,  at  a  premium  of  ^%,  brokerage  J^? 

,     8.  What  sum  must  be  invested  in  stocks  at  112,  paying  9^, 
to  obtain  a  yearly  income  of  $1260  ? 

Solution. — Since  $9  is  the  annual  income  on  1  share,  the  number  of 
shares  must  be  equal  $1260-^$9,  or  140  shares,  and  140  shares  at  $112  a 
share  amount  to  $15680,  the  required  investment. 

Find  the  investment  for  the  following: 
9.  Income  $2660,  stock  purchased  at  105J.  yielding  7^. 
10.  Income  $1800,  stock  purchased  at  109|,  yielding  12^. 

(236) 


STOCKS.  109 

11.  Income  $3900,  stock  purchased  at  92,  yielding  6^. 

12.  What  must  be  paid  for  stocks  yielding  1%  dividends,  that 
10%'  may  be  realized  annually  from  the  investment  ? 

Solution. — Since  $7,  the  annual  income  on  1  share,  must  be  10%  of 
the  cost  of  1  share,  yV  of  $7,  or  70  ct. ,  is  1  % .  Hence  100% ,  or  70  ct.  x  100 
=  $70,  is  the  amount  that  must  be  paid  for  the  stock. 

What  must  be  paid  for  stocks  yielding 

13.  b%  dividends* to  obtain  an  annual  income  of  8^  ? 

14.  1%  dividends  to  obtain  an  annual  income  of  12^  ? 

15.  9^  dividends  to  obtain  an  annual  income  of  7^  ? 

16.  How  much  currency  can  be  bought  for  1350  in  gold, 
when  the  latter  is  at  12;^  premium? 

Solution.— Since  $1  in  gold  is  worth  $1.12  in  currency,  $350  in  gold 
are  equal  $1.13  x  350  =  $392. 

How  much  currency  can  be  bought 

17.  For  $780  in  gold,  when  it  is  at  a  premium  of  9^? 

18.  For  1^396  in  gold,  when  it  is  at  a  premium  of  13-J^  ? 

19.  For  1520  in  gold,  when  it  is  at  a  premium  of  12^^? 

20.  How  much  is  1507.50  in  currency  worth  in  gold,  the 
latter  being  at  a  premium  of  12^%? 

Solution. — Since  $1  of  gold  is  equal  to  $1.12|  in  currency,  $507.50  in 
currency  must  be  worth  as  many  dollars  in  gold  as  $1.12|  is  contained 
times  in  $507.50,  which  is  $451. llj. 

How  much  gold  can  be  bought 

21.  For  $1053.17  currency,  when  gold  is  at  a  premium  of 

9i7;  ? 

22.  For  $317.47  currency,  when  gold  is  at  a  premium 
ofllf;^;? 

23.  For  $418.14  currency,  when  gold  is  at  a  premium 
of  13f  ^  ? 

24.  Bought  80  shares  in  Boston  and  Maine  Railroad,  at  a 
discount  of  2|^,  and  sold  the  same  at  an  advance  of  12^; 
what  did  I  gain  ?  Ans,  $1160. 

(237) 


110  S  USINES  S    ARITHMETIC, 

25.  An  agent  sells  415  barrels  of  flour,  at  $6  a  barrel,  com- 
mission b%,  and  invests  the  proceeds  in  'stocks  of  the  Suffolk 
Bank,  Boston,  at  17|-^  discount,  brokerage  i%;  how  many 
shares  did  he  buy  ? 

2G.  Bought  84  shares  in  Michigan  Southern  Eailroad,  at  1% 
discount,  and  sold  them  at  Q>\%  advance;  what  was  my  profit, 
the  brokerage  in  buying  and  selling  being  \  per  cent  ? 

27.  Bought  bonds  at  70^^,  bearing  4:\%  interest;  what  is  the 
rate  of  income  ?  Ans.  (j%. 

28.  I  invest  $2397.50  in  Empire  Iron  Foundry  stock,  whose 
shares,  worth  $50  each,  are  sold  at  ^43.50,  brokerage  J^;  what 
annual  income  shall  I  derive,  the  stock  yielding  1%  ? 

29.  0.  E.  Bonney  sold  $6000  Pacific  Eailroad  6's  at  107,  and 
with  a  part  of  the  proceeds  bought  St.  Lawrence  County 
bonds  at  90,  yielding  ^%  dividends  sufficient  to  give  an  annual 
income  of  $180  ;  how  much  has  he  left  ? 

30.  What  rate  of  income  can  be  derived  from  money  invested 
in  the  stock  of  a  company  paying  a  semi-annual  dividend  of 
b%,  purchased  at  84J^,  brokerage  ^%  ? 

31.  What  must  I  pay  for  bonds  yielding  4^^  annually,  that 
my  investment  may  pay  Q%  ? 

32.  What  must  be  paid  for  stocks  paying  5  per  cent,  that 
the  investment  may  return  %%  ? 

33.  How  much  more  is  $1400  gold  worth  than  11515  cur- 
rency, when  gold  is  112^? 

34.  A  man  bought  a  farm,  giving  a  note  for  $3400,  payable 
in  gold  in  5  years;  at  the  expiration  of  the  time  gold  was 
175/^:  what  did  his  farm  cost  in  currency? 

35.  I  invested  $785.40  of  currency  in  gold  when  it  waswortli 
115|;^ ;  what  amount  of  gold  did  I  purchase  ? 

36.  How  much  gold  at  a  jDremium  of  {)^%  can  be  purchased 
for  $876.90  currency  ?     For  $85.50  ?    For  $136.80  ? 

37.  What  is  the  difference  in  the  value  of  $800  in  gold  and 
$900  in  currency,  when  gold  is  at  a  premium  of  13|^^  ? 

(238) 


TAXES, 


111 


TAXES. 

541,  A  Tax  is  a  sum  of  money  assessed  upon  a  person  or 
property,  for  any  public  purpose. 

54^.  Property  is  of  two  kinds:  Real  Estate,  such  as 
liouses  and  lands ;  Personal  Property,  such  as  merchandise, 
cash,  furniture,  ships,  notes,  bonds,  mortgages,  etc. 

543.  Taxes  are  of  two  kinds :  Property  Tax,  which  is 
assessed  upon  taxable  property  according  to  its  estimated 
yalue ;  Poll  Tax,  which  is  a  sum  assessed  without  regard  to 
property  upon  each  male  citizen  liable  to  taxation. 

544.  An  Assessment  Eoll  is  a  schedule  or  list  which  con- 
tains the  names  and  the  taxable  value  of  the  property  of  all 
persons  subject  to  a  given  tax. 

545.  The  Hate  of  Property  Tax  is  the  rate  per  cent 
on  the  valuation  of  the  property. 

546.  An  Assessor  is  an  officer  appointed  to  prepare  the 
Assessment  Roll  and  apportion  to  each  person  his  tax. 

Method  of  A2>portloning  a  Tax, 

54*7.  1.  Tlie  Assessor  determines  by  a  personal  examination  the 
taxable  value  of  the  real  estate  and  personal  property  of  each  p6rson 
subject  to  tlie  tax,  and  fills  an  Assessment  Roll,  thus : 

ASSESSMENT  KOLL. 


NAMES. 

KEAL 
ESTATE. 

PERSONAL 
PROPERTY. 

TOTAL 
PROPERTY. 

POLLS. 

AMOUNT  OF 
TAX. 

L.  Henry, 
W.  Mann, 
P.  Duncan, 
R.  Storey, 

$6984 

8095 
9709 
0093 

$1863 
1983 
2300 
1975 

$8846 
10078 
12009 

8057 

1 

1 
1 
1 

$45.48 
51.64 
61.295 
41.585 

Totals. 

30880 

8120 

39000 

4 

$200. 

(239) 


112  BUSIIiUSS     ARITHMETIC, 

2.  If  the  amount  to  be  raised  on  this  Assessment  Roll  is  $195,  the 
collector's  fees  2|%,  and  the  poll  tax  $1.25  i)er  poll,  the  Assessor,  or 
party  authorized  to  do  so,  would  proceed  to  apportion  the  tax  thus  : 

(1.)  Since  the  collector  is  paid  2|%  of  the  whole  tax  for  collecting, 
the  $195  is  97 1%  of  the  assessment  which  must  be  made.  Hence, 
$195  -4- .97^  =  $200,  the  whole  tax. 

(2.)  Since  each  poll  pays  $1.25,  the  4  polls  will  pay  $5,  and  the 
amount  which  must  be  assessed  on  the  property  is  $200  —  $5  =  $195. 

(3.)  Since  $195  are  to  be  assessed  on  $39000,  the  whole  amount  of 
property,  the  rate  per  dollar  is  $195  -4-  $39000  =  .005,  or  5  mills. 

(4)  Multiplying  each  man's  taxable  property  by  .005  will  give  his 
property  tax,  to  which  we  add  the  poll  tax  ;  hence, 

J.  Henry's  tax     =     $8846  x  .005  +  $1.25  =  $45.48. 

.      W.Mann's  tax    =  $10078  x  .005  +  $1.25  =  $51.64. 

P.  Duncan's  tax  =  $12009  x  .005  +  $1.25  =:  $61,295. 
R.  Storey's  tax    =     $8067  x  .005  +  $1.25  =  $41,585. 

(5.)  These  results  are  now  inserted  in  the  blank  under  "Amount  of 
Tax,"  and  the  Assessment  Roll  is  thus  completed  ready  for  the 
collector. 

EXAMPIiES      FOR      PRACTICE. 

548,  Prepare  an  Assessment  Roll  ready  for  the  collector 
for  each  of  the  following : 

1.  Net  tax  to  be  raised  $1930,  collector's  fee  Z^%,  poll  tax 
$2.50  per  poll. 

Property  taxed. — A,  real  estate  $10800,  personal  property 
$3200 ;  B,  real  estate  $9600,  personal  property  $5200  ;  C, 
personal  property  $4200;  D,  real  estate  $12800,  personal 
property  $4000;  E,  real  estate  $20000,  personal  property 
$6200  ;  Polls  without  property  35. 

2.  Net  tax  to  be  assessed  $2387.50,  collector's  fee  A\%,  poll 
tax  $1.25  per  poll. 

Property  taxed. — A,  real  estate  $9700,  personal  property 
$5000;  B,  real  estate  $14600,  personal  property  $5400;  0, 
real  estate  $8900,  personal  property  $3100 ;  D,  real  estate 
$40000,  personal  property  $12000  ;  E,  real  estate  $21600,  per- 
sonal property  $3700 ;  Polls  without  property  11. 

(240) 


DUTIES    AND     CUSTOMS,  113 

DUTIES   OR  CUSTOMS. 

549.  Duties  or  Ctistoms  are  taxes  levied  by  the  gov- 
ernment upon  imported  goods. 

550.  A  Specific  Duty  is  a  certain  sum  imposed  upon 
an  article  without  regard  to  its  value. 

551.  An  Ad  Valorem  Duty  is  a  per  cent  assessed 
upon  the  value  of  an  article  in  the  country  from  which  it  is 
brought. 

552.  A  Tariff  is  a  schedule  giving  the  rates  of  duties 
fixed  by  law. 

553.  The  following  deductions  or  allowances  are  made 
before  computing  specific  duties : 

1.  Tare. — An  allowance  for  the  box,  cask,  bag,  etc.,  containing  the 
merchandise. 

2.  Leakage, — ^An  allowance  for  waste  of  liquors  imported  in  casks 
or  barrels. 

8.  Brealcage, — An  allowance  for  loss  of  liquors  imported  in  bottles 

EXAMPIiES     FOR     PRACTICES. 

554.  1.  "What  is  the  duty  on  420  boxes  of  raisins,  each 
containing  40  pounds,  bought  for  8  cents  a  pound,  at  20  per 
cent  ad  valorem  ? 

2.  Imported  21  barrels  of  wine,  each  containing  31  gallons ; 
2%  being  allowed  for  leakage,  what  is  the  duty  at  40  cents  per 
gallon  ? 

3.  A  merchant  imported  from  Havana  100  boxes  oranges 
@  $2.25  per  box ;  75  hogsheads  of  molasses,  each  containing 
G3  gal.,  @  23  cents  per  gal. ;  50  hogsheads  of  sugar,  each 
containing  340  lb.,  @  6  cents  per  lb.  The  duty  on  the  molasses 
was  25^,  on  the  sugar  30^,  and  on  the  oranges  20^.  What 
was  the  duty  on  the  whole  ? 

(341) 


114  BUSINES  S    ARITHMETIC, 

4.  What  is  the  duty  on  320  yards  of  cloth,  invoiced  at  11.15 
per  yard,  at  20%  ad  valorem  ? 

5.  At  12%  ad  valorem,  what  i»  the  duty  on  100  barrels 
of  kerosene,  invoiced  at  $.18  a  gallon,  2%  leakage  ? 

EEVIEW  AND   TEST   QUESTIONS. 

555.  1.  When  a  fraction  is  to  be  divided  by  a  fraction,  why 
can  the  factors  that  are  common  to  the  denominators  of  the 
dividend  and  divisor  be  cancelled  ? 

2.  How  does  moving  the  decimal  point  one  or  more  places 
to  the  left  or  right  affect  a  number,  and  why  ? 

3.  Show  that  multiplying  by  1000  and  subtracting  three 
times  the  multiplicand  from  the  product  is  the  same  as  multi- 
plying by  997  ? 

4.  Define  Base,  Percentage,  Amount  and  Difference. 

5.  When  the  amount  and  rate  per  cent  is  given  to  find  the 
base,  why  add  the  rate  expressed  decimally  to  1  and  divide  by 
the  result  ? 

6.  Eepresent  the  quantities  by  letters  and  write  a  formula 
for  solving  each  of  the  following  problems  (508). 

I.  Given,  the  Cost  and  the  Profit^  to  find  the  rate  per 
cent  profit. 
II.  Given,  the  rate  per  cent  profit  and  the  selling  price, 

to  find  the  buying  price, 
in.  Given,  the  amount  of  money  sent  to  an  agent  to 
purchase  goods  and  the  rate  per  cent  commission, 
to  find  the  amount  of  the  purchase. 
rV.  Given,  the  rate  at  which  stocks  can  be  purchased,  to 

find  how  much  can  be  secured  for  a  given  sum. 
V.  Given,  the  rate  at  which  stocks  can  be  purchased 
and  the  rate  per  cent  of  dividend,  to  find  the  rate 
per  cent  of  income  on  the  investment. 
VI.  Given,  the  premium  on  gold,  to  find  how  much  can 
be  purchased  for  a  given  sum  in  currency. 
(242) 


-•-•s^l  INTEREST  ifo^*^ 
^ ^^^ 

DEFINITIONS. 
55G.  Interest  is  a  sum  paid  for  the  use  of  money. 

Thus,  I  owe  Wm.  Henry  $200,  which  he  allows  me  to  use  for  one 
year  after  it  is  due.  At  the  end  of  the  year  I  pay  him  the  $200  and  $14 
for  its  use.     The  $14  is  called  the  Interest  and  the  $200  the  Principal. 

557.  Principal  is  a  sum  of  money  for  the  use  of  which 
interest  is  paid. 

558.  Rate  of  Interest  is  the  number  of  units  of  any 
denomination  of  money  paid  for  the  use  of  100  units  of  the 
same  denomination  for  one  year  or  some  j^ven  interval  of 
time. 

559.  The  Amount  is  the  sum  of  the  principal  and  interest. 

560.  Simxple  Interest  is  interest  which  falls  due  when 
the  principal  is  paid,  or  when  a  partial  payment  is  made. 

561.  Legal  Interest  is  interest  reckoned  at  the  rata 
per  cent  Jixcd  by  law. 

562.  Usury  is  interest  reckoned  at  a  higher  rate  than  is 
allowed  by  law. 

563.  The  following  table  gives  the  legal  rates  of  interest  in 

the  different  States. 

Where  two  rates  are  given,  any  rate  between  these  limits  is  allowed, 
if  specified  in  tenting.  When  no  rate  is  named  in  a  paper  involving 
interest,  the  legal  or  lowest  rate  is  always  understood. 

(243) 


116 


B  USINUSS    ARITHMETIC, 


STATES. 

BATE  5^. 

STATES. 

RATEjg. 

STATES. 

RATE  %. 

STATES. 

RATE^. 

Ala 

Ark 

Arizona.. 

Cal 

Conn 

Colo 

Dakota... 

Del 

D.C 

Flor 

Geo 

Idaho.... 

I 

10 
10 
7 
10 

I 

6 

8 
7 
10 

Any 
Any 
Any 

Any 
Any 

10 

Any 
10 

Ill 

Ind 

Iowa 

Kan 

Ken 

La 

Maine.... 

Md 

Mass 

Mich 

Minn.... 
Miss 

6 

6 
6 
7 
6 
5 
6 
6 
6 
7 
7 
6 

10 
10 
12 
10 
8 
Any 

Any 
10 
12 
10 

Mo 

Montana . 

N.H 

N.J 

N.Y 

N.C 

Neb 

Nevada  . 

Ohio 

Oregon,.. 

Penn 

R.I 

6 

10 

& 

7 
7 
6 
10 
10 
6 
10 
6 
6 

10 

8 

15 

An' 

8 

12 

7 

Any 

S.  C 

Tenn 

Texas.... 

Utah 

Vt 

Va 

W.Va.... 

W.  T 

Wis 

Wy 

7 
6 
8 

10 
6 
6 
6 

10 
7 

12 

Any 

10 

12 
Any 

12 

Any 
10 

The  legal  rate  for  England  and  France  is  5^ ;  for  Canada  and  Ireland,  Q%, 

564.  PROB.  I. — To  find  tlie  simple  interest  of  any- 
given  sum  for  one  or  more  years. 

1.  Find  the  interest  on  $384  for  5  years,  at  7^. 

Solution. — 1.  Since  the  interest  of  $100  for  one  year  is  |7,  the 
interest  of  $1  for  one  year  is  $.07.  Hence  the  interest  of  $1  for  5  years 
is  $.07x5  =  $.35. 

2.  Since  the  interest  of  $1  for  5  yr.  is  $  35,  the  interest  of  $384  for  the , 
same  time  must  be  384  times  $.35,  or  $134.40.     Hence  the  following 

565.  Rule.* — I.  Find  the  intei^est  of  $1  at  the  giiien 
rate  for  the  given  time,  and  multiply  this  result  hy  the 
numher  of  dollars  in  the  given  pHneipal. 

II.  To  -find  the  amount  add  the  interest  and  principal. 


EXAMPIi 


FOR     PRACTICE 


566.  Find  the  interest  on  the  following  orally: 


1.  $800  for  2  years  at  4^. 

2.  $1200  for  3  years  at  Z%, 

3.  $200  for  5  years  at  Q>%. 

4.  $600  for  4  years  at  b%. 

5.  $90  for  2  years  at  7^. 

6.  $70  for  4  years  at  %%. 


7.  $400  for  8  years  at  h%, 

8.  $100  for  12  years  at  9^. 

9.  e$600  for  7  years  at  10^ 

10.  $1000  for  5  years  at  S%, 

11.  $20  for  3  years  at  ^%. 

12.  $500  for  5  years  at  b% 
(244) 


SIMPLE    INTEREST,  117 

Find  the  interest  on  the  following : 

13.  1245.36  for  3  years  at  1%.     20.  $375.84  for  3  years  at  ^%. 

14.  $784.25  for  .9  years  at  4^.     21.  $293.50  for  6  years  at  ^%, 

15.  $836.95  for  2  years  at  ^%.    22.  $899.00  for  12  years  at  7f  ^. 

16.  $795.86  for  7  years  at  \%,    23.  $600.80  for  9  years  at  ^%. 

17.  $896.84  for  3J  years  at  2f^.    24.  $50.84  for  5  years  at  1^. 

1 8.  $28.95  for  1^  years  at  ^%.  25.  $95. 60  for  |  of  a  yr.  at  1^%, 

19.  $414.14  for  4  years  at  ^%,     26.  $262.62  for  6  years  at  ^%, 

METHOD  BY  ALIQUOT  PARTS. 

567.  Prob.  XL— To  find  the  interest  on  any  sum  at 
any  rate  for  years,  months^  and  days  by  aliquot  i>arts. 

1.  In  business  transactions  involving  interest,  30  days  are 
usually  considered  one  month,  and  12  months  1  year.  Hence 
the  interest  for  days  and  months  may  be  found  according  to 
(499),  by  regarding  the  time  as  a  compound  number;    thus, 

Find  the  interest  and  amount  of  $840  for  2  yr.  7  mo.  20  da., 

at  7^. 

$840    Principal. 
.07    Rate  of  Interest. 
6  mo.= J  of  1  yr.,  hence  2)  68.80    Interest  for  1  yr. 

2 

117.60    Interest  for  2  yr. 
1  mo. = J  of  6  mo.,  hence  6)  29.40  «        "  6  mo. 

15da.=  ^of  lmo.,hence2)    4.90  «        "  1  mo. 

5da.=^ofl5da.hence3)    2.45  "        "  15  da. 

81f        «        "  5  da. 
$155.1 6f        "        "  2  yr.  7  mo.  20  da. 
840.00    Principal. 
$995.16|  Amount  for  2  yr.  7  mo.  20  da 

568.  The  interest,  by  the  method  of  aliquot  parts,  is 
usually  found  by  finding  first  the  interest  of  $1  for  the  given 

(245) 


118  BUSINESS    ARITH3IETIC, 

time,  and  multiplying  the  given  principal  by  the  decimal 
expressing  the  interest  of  $1 ;  thus, 

Find  the  interest  of  1680  for  4  yr.  9  mo.  1 5  da.  at  8^. 

1.  We  first  find  tlie  interest  of  $1  for  tlie  given  time  thus  : 

8  ct.     =  Int.  of  $1  for  1  yr.,  8  ct.  x  4  ==  Int.  for  4  yr.    =  33  ct. 

6  mo.  =  ^  of  1  yr.,  hence,     -|-  of  8  ct.  =     "      «   6  mo.  —    4  ct. 

3  mo.  =  i  of  6  mo.,    "  ^  of  4  ct.  =     "      "3  mo.  =     2  ct. 

15  da.    =  i  of  3  mo.,    "  i  of  2  ct.  =    "     "  15  da.    =  .03|  m. 

Hence  the  interest  on  $1  for  4  yr.  9  mo.  15  da.  —  $.383^-. 

2.  The  decimal  .383^  expresses  the  part  of  $1  which  is  the  interest  of 
$1  for  the  given  time  at  the  given  rate.  Hence,  $680  x  .383|=$260.66|, 
is  the  interest  of  $680  for  4  yr.  9  mo.  15  da.,  at  8%  ;  hence  the  following 

569.  Rule. — I.  Find  by  aliquot  parts  the  interest  of 
$1  for  the  given  rate  and  time. 

II.  Multiply  the  principal  hy  the  decimal  expressing 
the  interest  for  $1,  and  the  product  will  he  the  required 
interest. 

III.  To  find  the  Amount,  add  the  interest  to  the 
principal. 

EXAMPIiES     FOR     PRACTICE. 

570.  Find  the  interest 

1.  Of  1284  for  3  yr.  8  mo.  12  da.  at  6% ;  at  8|^. 

2.  Of  1500.40  for  2  yr.  10  mo.  18  da.  at  7% ;  at  9^. 

3.  Of  $296.85  for  4  yr.  11  mo.  24  da.  at  S% ;  at  5^. 

4.  Of  $860  for  1  yr.  7  mo.  27  da.  at  4.-}% ;  at  7|-^. 

6.  Of  12940.75  for  3  yr.  11  mo.  17  da.  at  7%  ;  at  di%. 

6.  Find  the  amount  of  $250.70  for  2  yr.  28  da.  at  8%. 

7.  Find  the  amount  of  $38.90  for  3  yr.  13  da.  at  9%. 

8.  A  man  invested  $795  at  S%  for  4  yr.  8  mo.  13  da.  How 
much  was  the  amount  of  principal  and  interest  ? 

9.  Paid  a  debt  of  1384.60,  which  was  upon  interest  for 
11  mo.  16  da.  at  7%.    What  was  the  amount  of  the  payment? 

10.  Find  the  amount  of  $1000  for  9  yr.  11  mo.  29  da.  at  7%. 

(246) 


SIMPLE    INTEREST,  119 

METHOD    BY    SIX    PER    CENT. 

PJREP  ARJ^TOItY     STEPS. 

571.  Step  I. — To  find  the  interest  for  any  number  of 
months  at  6%. 

1.  Since  the  interest  of  $1  for  12  months,  or  1  yr.,  at  Q%y  is 
6  cents,  the  interest  for  two  months,  which  is  \  of  12  months, 
must  be  1  cent,  or  yj^  part  of  the  principal. 

2.  Since  the  interest  for  2  months  is  yott  of  the  principal, 
the  interest  for  any  number  of  months  will  be  as  many  times 
yj(j-  of  the  principal  as  2  is  contained  times  in  the  given  num- 
ber of  months.    Hence  the  following 

573.  EuLE. — /.  Move  the  decimal  point  in  the  pj^n- 
cipal  TWO  PLACES  to  the  left  (470),  prefixing  ciphers, 
if  necessary. 

II.  Multiply  this  result  hy  one-half  the  number  of 
months. 

Or,  Multiply  -^^-^  of  the  principal  hy  the  number  of 
months  and  divide  the  result  by  2. 

EXAMPLES     FOR      PRACTICE. 

573.  Find  the  interest  at  6^ 

1.  Of  $890  for  8  mo.  6.  Of  1398  for  1  yr.  6  mo. =18  mo. 

2.  Of  1973.50  for  10  mo.  7.  Of  $750  for  2  yr.  8  mo. 

3.  Of  $486.80  for  18  mo.  8.  Of  $186  for  4  yr.  2  mo. 

4.  Of  $364.40  for  7  mo.  9.  Of  $268  for  2  yr.  6  mo. 

5.  Of  $432.90  for  13  mo.  10.  Of  $873  for  1  yr.  11  mo. 

574.  Step  II. — To  find  the  interest  for  any  number  of 
days  at  6%. 

1.  Since  the  interest  of  $1  for  2  months  at  6%  is  1  cent,  the 
interest  for  1  month,  or  30  days,  must  be  |  cent  or  5  mills. 
And  since  6  days  are  \  of  30  days,  the  interest  for  6  days  must 
be  I  of  5  mills,  or  1  mill,  which  is  j^q^  of  the  principal. 

(247) 


120  BUSINESS    ARITH31ETIC. 

2.  Since  the  interest  for  6  days  is  -^^-^  of  the  principal,  the 
interest  for  any  number  of  days  will  be  as  many  times  yqwo  of 
the  principal  as  6  is  contained  times  in  the  given  number  of 
days.     Hence  the  following 

51i^,  KuLE. — /.  Move  the  deeimal  -point  in  the  prin- 
cipal THREE  PLACES  to  the  left  (470),  prefixing  ciphers, 
if  necessary. 

II.  Multiply  this  result  hy  one-sixth  the  numher  of 
days. 

Or,  Multiply  y^^^  of  the  principal  hy  the  number  of 
days  and  divide  the  result  by  6. 

EXAMPLES     FOR     PRACTICE. 

511G,  Find  the  interest  at  (j% 

1.  Of  $790  for  12  da.  6.  Of  1584  for  19  da. 

2.  Of  1384  for  24  da.  7.  Of  $730  for  22  da. 

3.  Of  1850  for  15  da.  8.  Of  $809  for  28  da. 

4.  Of  $935  for  27  da.  9.  Of  $396  for  17  da. 

5.  Of  $580  for  16  da.  10.  Of  $840  for  14  da. 

577.  Prob.  III. — To  find  the  interest  on  any  snm  at 
any  rate  for  years,  months,  and  days,  by  the  six  per 
cent  method. 

Find  the  interest  of  $542  for  4  years  9  months  17  days  at  8 
per  cent. 

Solution.— 1.  The  interest  of  $542  for  4  years  at  6^,  according  to 
(5<>4),  is  $542  X  .06  X  4  =  $130.08. 

2.  The  interest  for  9  months,  according  to  (571),  is  j^^  of  $543  or 
$5.42  multiplied  by  9,  and  this  product  divided  by  2  =  $24.39. 

3.  The  interest  for  17  days,  according  to  (574),  is  t^Vtt  ^^  $542  or 
$.542  multiplied  by  17,  and  this  product  divided  by  6  =  $1,535  +  . 

Hence  $130.08  +  $24.39  +  $1.54  =  $156.01,  the  interest  of  $542  for 
4  years  9  months  and  17  days. 

4.  Having  found  the  interest  of  $542  at  0%,  to  find  the  interest  at  8^ 
we  have  8%  -  6%  +  2%,  and  2%  is  i  of  6%.  Hence,  $156.01  +  ^  of 
$156.01  =  $208,013,  the  interest  of  $542  at  8%  for  4  yr.  9  mo.  17  da. 

(248) 


SIMPLE    INTEREST, 


121 


EXAMPLES      FOR      PRACTICE. 

518.  Find  the  interest  by  the  6%  method 

1.  Of  $384.96  for  2  yr.  8  mo.  12  da.  afc  G%;  at  9%;  at  S%. 

2.  Of  $890.70  for  4  yr.  10  mo.  15  da.  at  7% ;  at  10^;  at  4^. 

3.  Of  $280.60  for  11  mo.  27  da.  at  S%;  at  4:%;  at  7%. 

4.  Of  $480  for  2  yr.  7  mo.  15  da.  at  9%;  at  12^ ;  at  4J^. 
6.  Of  $890  for  9  mo.  13  da.  at  6^%;  at  8^;^ ;  at  9^%, 


METHOD    BY    DECIMALS. 

579,  In  this  method  the  time  is  regarded  as  a  compound 
number,  and  the  months  and  days  expressed  as  a  decimal  of  a 
year. 

When  the  principal  is  a  small  sum,  suflBcient  accuracy  will 
be  secured  by  carrying  the  decimal  to  three  places ;  but  when 
a  large  sum,  a  greater  number  of  decimal  places  should  be 
taken. 

580.  Peob.  IV. — To  find  the  interest  on  any  sum  at 
any  rate  for  years,  months,  and  days,  by  decimals. 

What  is  the  amount  of  $450  for  5  yr.  7  mo.  16  da.  at  6^? 

Explanation.— 1.  We  express,  accord- 
ing to  (389—15),  the  days  and  months  as 
a  decimal  of  a  year,  as  shown  in  (1). 

2.  We  find  the  interest  on  $450,  the 
given  principal,  for  1  year,  which  is  $27, 
as  shown  in  (3). 

3.  Since  $37  is  the  interest  on  $450  for 
1  year,  the  interest  for  5.637  years  is  5.637 
times  $37,  which  is  $151,929,  as  sliowu 
in(l). 

4.  The  amount  is  equal  to  the  principal 
plus  the  interest  (559);  hence,  $151.93 
+  $450  =  $601.93  is  the  amount.  Hence 
the  following 

(249) 


(1.) 

30  )  16        da. 

(2.) 
$450 

12  )    7.533  mo. 

.06 

5.627  yr. 

27 

$27.00 

39  389 
112  54 

$151,929    Interest. 
450 

$601.93      Amount 


1^  B  USIiYSSS    ARITHMETIC, 

581.  EuLE. — /.  To  find  the  interest,  multiply  the 
prineipal  by  the  rate,  and  this  product  by  the  tiine, 
expressed  in  years  and  decimals  of  a  year. 

II.  To  find  the  amount,  add  the  interest  to  the  prin- 
cipal. 

EXAMPLES     FOR      PRACTICE. 

582.  Find  the  interest  by  the  decimal  method 

1.  Of  $290  for  1  yr.  8  mo.  12  da.  at  b% ;  at  8^;  at  7%- 

2.  Of  $374.05  for  2  yr.  9  mo.  15  da.  at  0% ;  at  9%  ;  at  4^. 

3.  Of  $790.80  for  5  yr.  3  mo.  7  da.  at  7% ;  at  11^;  at  3%. 

4.  Of  $460.90  for  3  yr.  5  mo.  13  da.  at  Gi% ;  at  8^% ;  at  3|^. 
6.  Of  $700  for  11  mo.  27  da.  at  ^% ;  at  7i^;  at  2^%. 

6.  Of  $580.40  for  17  da.  at  6^%;  at  9^^ ;  at  5i^. 

7.  Of  $890  for  7  yr.  19  da.  at  6% ;  at  8^ ;  at  5^. 

EXACT  INTEREST. 

583.  In  the  foregoing  methods  of  reckoning  interest  the 
year  is  regarded  as  360  days,  which  is  5  days  less  than  a  com- 
mon year,  and  6  days  less  than  a  leap  year  ;  hence,  the  interest 
when  found  for  a  part  of  a  year  is  incorrect. 

Thus,  if  the  interest  of  $100  is  $7  for  a  common  year  or  365  days,  the 
interest  for  75  days  at  the  same  rate  must  be  //g  of  $7  ;  but  by  the  fore- 
going method  //^  of  $7  is  taken  as  the  interest,  -.vhich  is  too  great. 

Observe,  that  in  using  //j  instead  of  //;,,  the  denominator  is  dimin- 
ished ^f g^  =  j\  part  of  itself,  and  consequently  (330)  the  result  is  ^^ 
part  of  itself  too  great. 

Hence,  when  interest  is  calculated  by  the  foregoing  methods,  it  must 
be  diminished  by  -^^  of  itself  for  a  common  year,  and  for  like  reasons  g^ 
of  itself  for  a  leap  year. 

To  find  the  exact  interest  we  have  the  following  : 

584.  Rule. — /.  Find  the  interest  for  the  ^iven  num- 
ber of  years  (563). 

//.  Find  the  exact  number  of  days  in  the  given 
months  and  days,  and  take  such  a  part  of  the  interest 

(250) 


SIMPLE    INTEREST.  123 

of  the  principal  for  one  year,  as  the  whole  number  of 
days  is  of  365  days. 

Or,  Find  the  interest  for  the  given  months  and  days 
hy  either  of  the  foregoing  methods,  then  subtract  /j- 
part  of  itself  for  a  comjnon  year,  and  gV  f(^^  ^  leap 
year. 

III.  Add  the  result  to  the  interest  for  the  given  num- 
ber of  years. 

EXAMPLES     FOR     PRACTICE. 

5S^.  Find  the  exact  interest  by  both  rules 

1.  Of  1836  for  84  da.  at  6^.        5.  Of  $2300  for  7  da.  at  ^%. 

2.  Of  $2G0  for  55  da.  at  8^,        6.  Of  $120  for  133  da.  at  ^%. 

3.  Of  $690  for  25  da.  at  7^.        7.  Of  $380.50  for  93  da.  at  6f ^. 

4.  Of  $985  for  13  da.  at  9^.        8.  Of  $260.80  for  17  da.  at  12^. 

9.  Required  the  exact  interest  of  $385.75  at  7^,  from  Jan- 
uary 15,  1875,  to  Aug  23  following. 

10.  'What  is  the  difference  between  the  exact  interest  of 
$896  at  7^  from  January  11,  1872,  to  November  19, 1876,  and 
the  interest  reckoned  by  the  six  per  cent  method  ? 

11.  A  note  for  $360.80,  bearing  interest  at  8^,  was  given 
March  1st,  1873,  and  is  due  August  23,  1876.  How  much  will 
be  required  to  pay  the  note  when  due  ? 

12.  What  is  the  exact  interest  of  $586.90  from  March  13  to 
October  23  of  the  same  year,  at  1%  ? 

586.  Prob.  V. — To  find  the  principal  when  the  inter- 
est, time,  and  rate  are  g^iven. 

Obserneyihaii  the  interest  of  any  principal  for  a  given  time  at  a  given 
rate,  is  the  interest  of  $1  taken  (503)  as  many  times  as  there  are 
dollars  in  the  principal  ;  hence,  the  following 

58?.  Rule. — Divide  the  given  interest  by  the  interest 
of  $1  for  the  given  time  at  the  given  rate. 

(351) 


124  BUSINUSS    ARITHMETIC. 


EXAMPLES     FOR     PRACTICE. 

588.  1.  What  sum  of  money  will  gain  $110.25  in  3  yr. 
9  mo.  at  11%  ? 

Solution. — Tlie  interest  of  $1  for  3  yr.  9  mo.  at  7%,  is  $.2625.  Now 
fsince  $.2625  is  the  interest  of  $1  for  the  given  time  at  the  given  rate, 
$110.25  is  the  interest  of  as  many  dollars  for  the  same  time  and  rate  as 
$.2625  is  contained  times  in  $110.25.  Hence  $110.25  ^  .2625  =  $420, 
the  required  principal. 

What  principal  or  sum  of  money 

2.  Will  gain  $95,456  in  3  yr.  8  mo.  25  da.  at  7%? 

3.  Will  gain  $63,488  in  2  yr.  9  mo.  16  da.  at  S%  ? 

4.  Will  gain  $106,611  in  3  yr.  6  mo.  18  da.  at  6^%  ? 
6.  Will  gain  $235,609  in  4  yr.  7  mo.  24  da  at  9%  ? 

6.  Will  gain  $30,636  in  1  yr.  9  mo.  18  da.  at  5}^? 

7.  Will  gain  $74,221  in  2  yr.  3  mo.  9  da.  at  '7^%  ? 

589.  Prob.  VI.— To  find  the  principal  when  the 
amount,  time,  and  rate  are  given. 

Observe,  that  the  amount  is  the  principal  plus  the  interest,  and  that 
the  interest  contains  the  interest  (564)  of  $1  as  many  times  as  there  are 
dollars  in  the  principal ;  consequently  the  amount  must  contain  (500) 
$1  plus  the  interest  of  $1  for  the  given  time  at  the  given  rate  as  many 
times  as  there  are  dollars  in  the  principal ;  hence,  the  following 

590.  EuLE. — Divide  the  amount  hy  the  amount  of  $1 
for  the  given  time  at  the  given  rate. 


EXAMPIiES    FOR     PRACTICB. 

591,  1.  What  sum  of  money  will  amount  to  $290.50  in 
2  yr.  8  mo.  12  da.  at  6^  ? 

Solution. —The  amount  of  $1  for  2  yr.  8  mo.  12  da.  at  6%  is  $1,162. 
Now  since  $1,162  is  the  amount  of  $1  for  the  given  time  at  the  given 
rate,  $290.50  is  the  amount  of  as  many  dollars  as  $1,162  is  contained 
times  in  it    Hence,  $290.50  -f-  $1 162  —  $250,  is  the  required  principal. 

(252) 


SIMPLE    INTEREST.  125 

2.  What  principal  will  amount  to  1310.60  in  3  yr.  5  mo. 
9  da.  at  b%  ? 

3.  What  is  the  interest  for  1  yr.  7  mo.  13  da.  on  a  sum  of 
money  which  in  this  time  amounts  to  $487.65,  at  7%  ? 

4.  What  sum  of  money  at  10^  will  amount  to  $436.02  in 
4  yr.  8  mo.  23  da. 

5.  At  S%  a  certain  principal  in  2  yr.  9  mo.  6  da.  amounted 
to  $699.82.     Find  the  principal  and  the  interest. 

592.  Prob.  VII.— To  find  the  rate  when  the  princi- 
pal, interest  and  time  are  given. 

Observe,  that  the  given  interest  must  be  as  many  times  1%  oi  the 
given  principal  for  the  given  time  as  there  are  units  in  the  rate  ;  hence 
the  following 

593.  Rule. — Divide  the  given  interest  hy  the  interest 
of  the  given  principal  for  the  given  time  at  1  per  cent. 

SXAMPIiES     FOR     PRACTICE. 

594.  1.  At  what  rate  will  $260  gain  $45.50  in  2 yr.  6  mo.? 

Solution.— The  interest  of  $260  for  2  yr.  6  mo.  at  1%  is  $6.50.  Now 
Bince  $6.50  is  1  %  of  $260  for  the  given  time,  $45.50  is  as  many  per  cent 
as  $6.50  is  contained  times  in  $45.50 ;  hence,  $45.50  -i-  $6.50  =  7,  is  the 
required  rate. 

At  what  rate  per  cent 

2.  Will  $732  gain  $99,674  in  2  yr.  3  mo.  7  da.  ? 

3.  AVill  $524  gain  $206.63  in  5  yr.  7  mo.  18  da.  ? 

4.  Will  $873  gain  $132.89  in  1  yr.  10  mo.  25  da.? 

5.  Will  $395.80  gain  $53,873  in  2  yr.  8  mo.  20  da.? 

6.  Will  $908.50  gain  $325,422  in  4  yr.  2  mo.  17  da.  ? 

7.  A  man  purchased  a  house  for  $3486,  which  rents  for 
$418.32.  What  rate  per  cent  does  he  make  on  the  invest- 
ment ? 

8.  Which  is  the  better  investment  and  what  rate  per  cent 

(253) 


126  BUSINESS    ARITHMETIC. 

per  annum,  ^4360  which  yields  in  5  years  11635,  or  $3860 
which  yields  in  9  years  12692.45  ? 

9.  At  what  rate  per  cent  per  annum  will  a  sum  of  money 
double  itself  in  7  years  ? 

Solution.— Since  in  100  years  at  1  %  any  sum  doubles  itself,  to  double 
itself  in  7  years  the  rate  per  cent  must  be  as  many  times  Ifo  as  T  is  con- 
tained times  in  100,  which  is  14f .     Hence,  etc. 

]  0.  At  what  rate  per  cent  per  annum  will  any  sum  double 
itself  in  4,  8,  9,  12,  and  25  years  respectively  ? 

11.  At  what  rate  per  cent  per  annum  will  any  sum  triple  or 
quadruple  itself  in  6,  9,  14,  and  18  years  respectively  ? 

12.  Invested  $3648  in  a  business  that  yields  11659.84  in  5 
years.  What  per  cent  annual  interest  did  I  receive  on  my 
investment  ? 

595.  Prob.  VIII. — To  find  the  time  when  the  princi- 
pal, interest,  and  rate  are  given. 

Observe,  that  the  interest  is  found  (580)  by  multiplying  the  interest 
of  the  given  principal  for  1  year  at  the  given  rate  by  the  time  expressed 
in  years  ;  hence  the  following 

596.  EuLE. — /.  Divide  the  given  interest  hy  the  inter- 
est of  the  given  principal  for  1  year  at  the  given  rate. 

II.  Reduce  (579),  when  called  for,  fractions  of  a 
year  to  months  and  days, 

BXAMPIiBS     FOR     PRACTICS. 

597.  1.  In  what  time  will  S350  gain  $63  at  8,^  ? 
Solution.— The  interest  of  $350  for  1  yr.  at  8%  is  $28.    Now  since 

$28  is  the  interest  of  $350  at  8%  for  1  year,  it  will  take  as  many  years 
to  gain  $63  as  $28  is  contained  times  in  $63  ;  hence  $63  h-  $28  =  2\  jt., 
or  2  yr.  3  mo.,  the  required  time. 

In  what  time  will 

2.  $80  gain  $36  at  7i%?  6.  $477  gain  $152.64  at  12^? 

3.  $460  gain  $80.50  at  5%  ?  6.  $600  gain  $301,392  at  7^%? 

4.  $260  gain  $98.80  at  8%  ?  7.  $385  gain  $214.72  at  S^^  ? 

(254) 


COMPOUND     INTEREST.  127 

8.  My  total  gain  on  an  investment  of  $860  at  1%  per  annum, 
is  $455.70.     How  long  has  the  investment  been  made  ? 

9.  How  long  will  it  take  any  sum  of  money  to  double  itself 
at  1%  per  annum  ? 

Solution. — At  100%  any  sum  will  double  itself  in  1  year  ;  hence  to 
double  itself  at  7/^  it  will  require  as  many  years  as  7%  is  contained 
times  in  100%,  which  is  14f.    Hence,  etc. 

Observe,  that  to  find  how  long  it  will  take  to  triple,  quadruple,  etc.,  any 
sum,  we  must  take  200%,  300%,  etc. 

10.  How  long  will  it  take  any  sum  of  money  at  b%,  8%,  6  J^, 
or  9%  per  annum  to  double  itself?    To  triple  itself,  etc.  ? 

11.  At  7%  the  interest  of  $480  is  equal  to  5  times  the  prin- 
cipal,    How  long  has  the  money  been  on  interest  ? 

COMPOUND   INTEREST. 

598.  Compound  Interest  is  interest  upon  principal 
and  interest  united,  at  given  intervals  of  time. 

Observe,  that  the  interest  may  be  made  a  part  of  the  principal,  or 
compounded  at  any  interval  of  time  agreed  upon  ;  as,  annually,  semi-an- 
nually, quarterly,  etc. 

599.  Prob.  IX. — To  find  the  compound  interest  on 
any  sum  for  any  given  time. 

Find  the  compound  interest  of  $850  for  2  yr.  6  mo.  at  6^. 

$850  Prin.  for  Ist  yr.           EXPLANATION.— Since  at  6%  the  amount 

IQQ  is  1.06  of  the  principal,  we  multiply  $850, 

7 the  principal  for  the  first  year,  by  1.06,  giv- 

$901  Prin.  for  2d  yr.      i^g  ^901^  the  amount  at  the  end  of  the  first 

1.06  year,  which  forms  the  principal  for   the 

^955.00     Prin.  for  6  mo.       *^^^^  >^^^^'     ^^  *^^  ^^^®  manner  we  find 
^  „^  $955.06,    the  amount  at  the   end   of    the 

1_L  second  year  which  forms  the  principal  for 

$983.71      Total  amount.       the  6  months. 

1850  Given  Prin.  2.  Since  6%   for  one  year  is  dfo  for  6 

months,  we  multiply  $955.06,  the  principal 

$133.71  Compound  Int.  fo^  ^^^^  g  months,  by  1.03,  which  gives  the 
total  amount  at  the  end  of  the  2  years  6  months. 

(255) 


12S  BUSINUSS    ARITHMETIC. 

3.  From  the  total  amount  we  subtract  $850,  the  given  principal,  which 
gives  $133.71,  the  compound  interest  of  $850  for  2  years  6  months  at  6^. 
Hence  the  following 

600.  EuLE. — I.  Find  the  amount  of  the  principal  for 

the  first  interval  of  time  at  the  end  of  which  interest  is 
due,  and  make  it  the  principal  for  the  second  interval. 

II.  Find  the  amount  of  this  principal  for  the  second 
interval  of  time,  and  so  continue  for  each  successive 
interval  and  fraction  of  an  interval,  if  any. 

III.  Subtract  the  given  principal  from  the  last 
amount  and  the  remainder  will  he  the  compound 
interest, 

CXAMPLCS     FOR     PRACTICE. 

601.  1.  What  is  the  compound  interest  of  $650  for  3  years, 
at  7^,  payable  annually? 

2.  Find  the  amount  of  $870  for  2  years  at  6^  compound 
interest. 

3.  Find  tlie  compound  interest  of  1380.80  for  1  year  at  8^, 
interest  payable  quarterly. 

4.  What  is  the  amount  of  $1500  for  2  years  9  months  at  8^ 
compound  interest,  payable  annually  ? 

5.  What  is  the  amount  of  $600  for  1  year  9  months  at  h% 
compound  interest,  payable  quarterly  ? 

6.  What  is  the  difference  in  the  simple  interest  and  com- 
pound interest  of  $480  for  4  yr.  and  6  mo.  at  7%'  ? 

7.  What  is  the  annual  income  from  an  investment  of  $2860 
at  7^  compound  interest,  payable  quarterly  ? 

8.  What  will  be  the  compound  interest  at  the  end  of  2  yr. 
5  mo.  on  a  note  for  $600  at  7^,  payable  semi-annually  ? 

9.  A  man  invests  $3750  for  3  years  at  7^  compound  interest, 
payable  semi-annually,  and  the  same  amount  for  the  same 
time  at  '^\%  simple  interest.  Which  will  yield  the  greater 
amount  of  interest  at  the  end  of  the  time,  and  how  much  ? 

(256) 


INTEREST     TABLES, 


120 


INTEREST  TABLES. 

602.  Interest,  both  simple  and  compound,  is  now  almost 
invariably  reckoned  by  means  of  tables,  which  give  the  interest 
or  amount  of  SI  at  different  rates  for  years,  months,  and  days. 
The  following  illustrate  the  nature  and  use  of  such  tables. 

Table  showing  the  simple  interest  of  $1  at  6,  7,  and  8% 
for  years,  months,  and  days. 


Years, 

1 

6%. 

7/.. 

.08 

Years^ 

4 

6^. 

7/. 

8%. 

.06 

.07 

.34 

.28 

.32 

3 

.12 

.14 

.16 

5 

.30 

.35 

.40 

3 

.18 

21 

.24 

6 

.36 

.42 

.48 

Months, 

1 

Months. 

7 

.005 

.00583 

.00666 

.035 

.04083 

.04666 

2 

.01 

.01166 

.01833  1 

8 

.04 

.04666 

.05333 

3 

.015 

.01750 

.02000! 

9 

.045 

.05250 

.06000 

4 

.02 

.03333 

.02660 

10 

.05 

.05833 

.06666 

5 

.025 

.02916 

.03333 

11 

.055 

.06416 

.07333 

6 

Days, 

1 

.03 

.03500 

.04000 

Days, 

16 

.00016 

00019 

.00032  1 

— ■ 

.00366 

.00311 

.00355 

2 

.00033 

.00038 

.00044 

17 

.00283 

.00330 

.00377 

3 

.00050 

.00058 

.00066 

18 

.00300 

.00350 

.00400 

4 

.00060 

.00077 

.00088 

19 

.00316 

.00369 

.C0422 

5 

.00083 

.00007 

.00111 

20 

.00333 

.00388 

.00444 

6 

.00100 

.00116 

.00133  i 

31 

.00350 

.00408 

.00160 

7 

.00116 

.00136 

.00155 

32 

.00366 

.00427 

.00488 

8 

.00133 

.00155 

.00177 

33 

.00383 

.00447 

.00511 

9 

.00150 

.00175 

.00200 

24 

.00400 

.00466 

.00533 

10 

.00166 

.00194 

.00223 

25 

.00416 

.00486 

.00555 

11 

.00183 

.00213 

.00344 

36 

.00433 

.00505 

.00577 

12 

.00200 

.00333 

.00266 

37 

.00450 

.00525 

.00600 

13 

.00216 

.00252 

.00388 

38 

.00466 

.00544 

.00633 

14 

.00233 

.00272 

.00311 

39 

.00483 

.00563 

.00644 

15 

.00250 

.00291 

.00333 

(257) 


130 


B  USINES  S     ARITHMETIC, 


Method  of  using  the  Simple  Interest  Table, 

603.  Find  the  interest  of  $250  for  5  yr.  9  mo.  18  da.  at  7;^. 

1.  We  find  the  interest  of  %\\       j  f^^^  ^^*"'^^<^  ^^  *^^1^  ^°^  ^  y^' 

for,.,^^t^  \-\^^  ^^    ;;  ;;   :;^- 

Interest  of  $1  for  5  yr.  9  mo.  18  da  is  .406     of  $1. 

2.  Since  the  interest  of  $1  for  5  yr.  9  mo.  18  da.  is  .406  of  $1,  the 
interest  of  $350  for  the  same  time  is  .406  of  $350. 

Hence,  $350  x  .406  =  $101.50,  the  required  interest. 


EXAMPLES      FOR      PRACTICE. 

604.  Find  by  using  the  table  the  interest,  at  ^%,  of 

1.  $860  for  3  yr.  7  mo.  23  da.  4.  $325.86  for  5  yr.  13  da. 

2.  $438  for  5  yr.  11  mo.  19  da.        5.  $796.50  for  11  mo.  28  da. 

3.  $283  for  6  yr.  8  mo.  27  da.  6.  $395.75  for  3  yr.  7  mo. 

Table  sliowing  tlie  amount  of  $1  at  6,  7,  and  8%  compound 
interest  from  1  to  12  years. 


YRS. 

6%. 

7%. 

8%. 

YRS. 

7 

6%. 

7%. 

8f., 

1 

1.060000 

1.070000 

1.080000 

1.503630 

1.605781 

1.713824 

2 

1.123000 

1.144900 

1.166400  1 

8 

1.593848 

1.718180 

1.850930 

3 

1.191016 

1.235043 

1.269712 

9 

1.089479 

1.838459 

1.999005 

4 

1.362477 

1.310796 

1.360489 

10 

1.790848 

1.967151 

3.158935 

5 

1.338226 

1.402552 

1.469328 

11 

1.898299 

2.104852 

3.331639 

6 

1.418519 

1.500730 

1.586874 

13 

2.013197 

2.252192 

3.518170 

3Iethod  of  using  the  Compound  Interest  Table, 

605.  Find  the  compound  interest  of  $2800  for  7  years 
at  G^. 

1.  The  amount  of  $1  for  7  years  at  6%  in  the  tahle  is  1.50303. 

3.  Since  the  amount  of  $1  for  7  years  is  1 .50363,  the  amount  of  $3800 
for  the  same  time  must  be  3800  times  $1.50363  =  $1.50303  x  2800  r:^ 
$4310.164.    Hence,  $4210. 164 -$3800  =  $1410. 161,  the  required  interest. 

(258) 


ANNUAL     INTEREST,  131 

BXAMPIiES     FOR      PRACTICE. 

606.  Find  by  using  the  table  the  compound  interest  of 

1.  $500  for  9  yr.  at  1%,  8.  1384.50  for  8  yr.  at  Q%, 

2.  82000  for  5  yr.  at  8^.  9.  8400  for  4  yr.  1  mo.  at  1%. 

3.  8870  for  11  yr.  at  Q%.  10.  $900  for  6  yr.  3  mo.  at  8^. 

4.  83800  for  7  yr.  at  8^.  11.  8690  for  12  yr.  8  mo.  at  G^. 

5.  82500  for  3  J  yr.  at  6^.  12.  $4000  for  9  yr.  2  mo.  at  1%. 

6.  8640  for  4J  yr.  at  8^.  13.  83900  for  4  yr.  3  mo.  at  Q%, 

7.  $285  for  9J  yr.  at  !%•  14.  8600  for  11  yr.  G  mo.  at  8^. 

ANNUAL  INTEREST. 

60T.  Annual  Interest  is  simple  interest  on  the  prin- 
cipal, and  each  year's  interest  remaining  unpaid. 

Annual  interest  is  allowed  on  promissory  notes  and  other 
contracts  which  contain  the  words,  "  interest  payable  annually 
if  the  interest  remains  unpaid." 

608.  Prob.  X. — To  find  the  annual  interest  on  a 
promissory  note  or  contract. 

What  is  the  interest  on  a  note  for  8600  at  7^  at  the  end  of 
3  yr.  6  mo.,  interest  payable  annually,  but  remaining  unpaid. 

Solution. — 1.  At  7^  the  payment  of  interest  on  $600  due  at  the  end 
of  each  year  is  $42,  and  the  simple  interest  for  3  yr.  6  mo,  is  $147. 

2.  The  first  payment  of  $42  of  interest  is  due  at  the  end  of  the  first 
year  and  must  bear  simple  interest  for  2  yr.  6  mo.  The  second  payment 
is  due  at  the  end  of  the  second  year  and  must  bear  simple  interest  for 
1  y r.  6  mo, ,  and  the  third  payment  being  due  at  the  end  of  the  third  year 
must  bear  interest  for  6  mo. 

Hence,  there  is  simple  interest  on  $42  for  2  yr.  6  mo.  +  1  yr.  6  mo.  + 
6  mo.  =  4  yr.  6  mo. ,  and  the  interest  of  the  $42  for  this  time  at  7  %  is 
$13.23. 

3.  The  simple  interest  on  $600  being  $147,  and  the  simple  interest  on 
the  interest  remaining  unpaid  being  $13.23,  the  total  interest  on  the  note 
at  the  end  of  the  given  time  is  $160.23. 

(259) 


Id2  BUSINESS    ARITHMETIC, 

EXAMPIiES     FOR     PRACTICE. 

609.  1.  How  much  interest  is  due  at  the  end  of  4  yr.  9  mo. 
on  a  note  for  $460  at  G%,  interest  payable  annually,  but 
remaining  unpaid  ? 

2.  Wilbur  H.  Reynolds  has  J.  G.  MacVicar's  note  dated 
July  29, 1876,  for  $800,  interest  payable  annually ;  what  will 
be  due  November  29,  1880,  at  7^  ? 

3.  Find  the  amount  of  $780  at  11%  annual  interest  for  5i  yr. 

4.  What  is  the  difference  between  the  annual  interest  and 
the  compound  interest  of  $1800  for  7  yr.  at  7^  ? 

5.  What  is  the  annual  interest  of  $830  for  4  yr.  9  mo.  at  S%  ? 

6.  What  is  the  difference  in  the  simple,  annual,  and  com- 
pound interest  of  $790  for  5  years  at  S%  ? 


PARTIAL  PAYMENTS. 

610.  A  Promissory  Note  is  a  written  promise  to  pay 
a  sum  of  money  at  a  specified  time  or  on  demand. 

The  Face  of  a  note  is  the  sum  of  money  made  payable  by  it. 

The  Maker  or  Drawer  of  a  note  is  the  person  who  signs  the  note. 

The  Payee  is  the  person  to  whom  or  to  whose  order  the  money  is 
paid. 

An  Indorser  is  a  person  who  signs  his  name  on  the  back  of  the  note, 
and  thus  makes  himself  responsible  for  its  payment. 

611.  A  Negotiable  JSFote  is  a  note  made  payable  to  the 
bearer,  or  to  some  person's  order. 

When  a  note  is  so  written  it  can  be  bought  and  sold  in  the  same 
manner  as  any  other  property. 

612.  A  Partial  Payment  is  a  payment  in  part  of  a 
note,  bond,  or  other  obligation. 

613.  An  Indorsement  is  a  written  acknowledgment  of 
a  partial  payment,  placed  on  the  back  of  a  note,  bond,  etc, 
stating  the  time  and  amount  of  the  same. 

(260) 


PARTIAL     PAYMENTS,  133 

MERCANTILE   RULE. 

614.  The  method  of  reckoning  partial  payments  known  as 
the  Mercantile  Rule  is  very  commonly  used  in  computing 
interest  on  notes  and  amounts  running  for  a  year  or  less.  The 
rule  is  as  follows : 

6J  5.  Rule. — I.  Find  the  amount  of  the  note  or  debt 
from  the  time  it  begins  to  bear  interest,  and  of  each 
paynieiit  until  the  date  of  settlement. 

II.  Subtract  the  sum  of  the  amounts  of  payments 
from  the  amount  of  the  note  or  debt;  the  remainder  will 
be  the  balance  due. 

Observe,  that  an  accurate  application  of  the  rule  requires  that  the 
exact  interest  should  be  found  according  to  (683). 

EXAMPLES     FOR     PRACTICE, 

616.  .  1.  $900.  Potsdam,  N.  Y.,  Sept.  M,  1876. 

On  demand  I  promise  to  pay  Henry  Watkins,  or  order, 
nine  hundred  dollars  with  interest,  value  received. 

Warrei^  Mankt. 

Indorsed  as  follows:  Oct.  18th,  1876,  $150;  Dec.  22d,  1876, 
$200 ;  March  15th,  1877,  $300.  What  is  due  on  the  note 
July  19th,  1877  ? 

2.  A  note  for  $600  bearing  interest  at  8^  from  July  1st, 
1874,  was  paid  May  16th,  1875.  The  indorsements  were: 
July  12th,  1874,  $185;  Sept.  15,1874,  $76;  Jan.  13,  1875, 
$230  ;  and  March  2,  1875,  $115.  What  was  due  on  the  note 
at  the  time  of  payment  ? 

3.  An  account  amounting  to  $485  was  due  Sept.  3, 1875, 
and  was  not  settled  until  Aug.  15, 1876.  The  payments  made 
upon  it  were:  $125,  Dec.  4,  1875;  $84,  Jan.  17,  1876;  $95, 
June  23,  1876.  What  was  due  at  the  time  of  settlement, 
allowing  interest  at  1%  ? 

(361) 


134  B  USINESS    ARITHMETIC, 

4.  1250.  Ogdensburg,  N.  Y.,  Mardi  25,  1876. 

Ninety-eight  days  after  date  I  promise  to  pay  E.  D.  Brooks, 
or  order,  two  hundred  fifty  dollars  with  interest,  value  received. 

Silas  Jones. 

Indorsements :  187,  April  12, 1876 ;  $48,  May  9.  What  is  to 
pay  when  the  note  is  due  ? 

UNITED  STATES  EULE. 

617.  The  United  States  courts  have  adopted  the  following 
for  reckoning  the  interest  on  partial  payments : 

618.  Rule. — I.  Find  the  aviount  of  the  given  prin- 
cipal to  the  time  of  the  first  payment ;  if  the  payment 
equals  or  exceeds  the  interest  then  due,  subtract  it  from 
the  avvount  obtained  and  regard  the  rem^ainder  as  the 
new  principal. 

II.  If  the  payment  is  less  than  the  interest  due,  find 
the  amount  of  the  given  pHncipal  to  a  time  when  the 
sum  of  the  payments  equals  or  exceeds  the  interest  then 
due,  and  subtract  the  sum  of  the  payments  from  this 
amount,  and  regard  the  remainder  as  the  new  prin- 
cipal. 

III.  Proceed  with  this  new  pHncipal  and  with  each 
succeeding  principal  in  the  same  manner. 

619.  The  method  of  applying  the  above  rule  will  be  seen 
from  the  following  example  : 

1.  A  note  for  ^900,  dated  Syracuse,  Jan.  5th,  1876,  and 
paid  Dec.  20th,  1876,  had  endorsed  upon  it  the  following 
payments:  Feb.  23d,  1876,  $40;  April  26th,  $6;  July  19th, 
1876,  $70.  How  much  was  the  payment  Dec.  20th,  1876, 
interest  at  7^  ? 

(262) 


PARTIAL     PAY31ENTS,  135 

SOLUTION. 

First  Step. 

1.  The  first  principal  is  the  face  of  the  note  .......     $900 

2.  We  find  the  interest  from  the  date  of  the  note  to  the  first 

payment,  Feb.  23,  1876  (49  da.),  at  7% 8.43 

Amount $908.43 

5.  The  first  payment,  $40,  being  greater  than  the  interest  then 

due,  is  subtracted  from  the  amount 40.00 

Second  prindpcU   ....    $868.43 

Second  Step, 

1.  The  second  principal  is  the  remainder  after  subtracting  the 

first  payment  from  the  amount  at  that  date      ....     $868.43 

2.  The  interest  on  $868.43,  from  Feb.  23  to  Apr.  26, 

1876  (63  da.),  is $10,463 

3.  This  interest  being  greater  than  the  second  pay- 

ment ($6),  we  find  the  interest  on  $868.55 
from   April  26  to  July  19,  1876,  (84  da.), 

which  is 13.951 

Interest  from  first  to  third  payment   .       $24,414        24.414 
Amount $892,844 

4.  Tlie  sum  of  the  second  and  third  payments  being  greater 

than  the  interest  due,  we  subtract  it  from  the  amount  .         76 

Third  principal    ....     $816,844 

Third  Step. 

We  find  the  interest  on  $816,844,  from  July  19  to  Dec.  20, 

1876  (154  da.),  which  is 24.057 

Payment  due  Dec.  20,  1876 $840.90 

In  the  above  example,  the  interest  has  been  reckoned 
according  to  (583) ;  in  the  following,  360  days  have  been 
regarded  as  a  year. 

EXAMPLES     FOR     PRACTICE. 

2.  A  note  for  $1630  at  S%  interest  was  dated  March  18, 
1872,  and  was  paid  Aug.  13,  1875.  The  following  sums  were 
endorsed  upon  it:  $160,  Feb.  12,  1873;  $48,  March  7, 18 H; 
and  $350,  Aug.  25,  1874.    How  much  was  paid  Aug.  13? 

(263) 


136  BUSINESS    ARITHMETIC. 

3.  A  mortgage  for  ^^3500  was  dated  Aug.  24,  1873.  It  had 
endorsed  upon  it  the  following  payments :  May  17,  1874,  $89  ; 
Sept.  12,  1874,  $635;  March  4,  1875,  $420.  How  much  was 
due  upon  it  Feb.  9, 1876,  interest  at  11%  ? 

4.  What  was  the  last  payment  on  a  note  for  $1000  at  7^, 
which  was  dated  Jan.  7,  1876,  and  paid  Dec.  26,  1876, 
endorsed  as  follows :  April  12,  1876,  $16  ;  July  10, 1876,  $250 ; 
and  Oct.  26,  1876,  $370  ? 

'  5.  A  mortgage  for  $4600,  dated  Leavenworth,  Kansas, 
Sept.  25,  1871,  had  endorsed  upon  it:  $400,  June  23,  1872; 
$125,  Aug.  3, 1873 ;  $580,  May  7, 1874 ;  and  $86,  Mar.  5,  1875. 
How  much  was  due  upon  it  Sept.  25, 1875,  interest  at  11^  ? 

DISCOUNT. 

620.  Discount  is  a  deduction  made  for  any  reason  from 
an  account,  debt,  price  of  goods,  and  the  like,  or  for  the  interest 
of  money  advanced  upon  a  bill  or  note  due  at  a  future  date. 

631.  The  Present  Worth  of  a  note,  debt,  or  other  obliga- 
tion, payable  at  a  future  time  without  interest,  is  such  a  sum 
as,  being  placed  at  interest  at  a  legal  rate,  will  amount  to  the 
given  sum  when  it  becomes  due. 

632.  True  Discount  is  a  difference  between  any  sum 
of  money  payable  at  a  future  time  and  its  present  ivorth. 

633.  Prob.  XL — To  find  the  present  worth  of  any 
sum. 

Find  the  present  worth  of  a  debt  of  $890,  due  in  2  yr.  6  mo. 
without  interest,  allowing  %%  discount. 

Solution.— Since  $1  placed  at  interest  for  2  yr.  6  mo.  at  8%  amounts 
to  $1.20,  the  present  worth  (021)  of  $1.20,  due  in  3  yr.6  mo.,  is  $1. 
Hence  the  present  worth  of  $890,  which  is  due  ■^^ithout  interest  in  2  yr. 
6  mo.,  must  contain  as  many  dollars  as  $1.20  is  contained  times  in 
$890  =  $741.66. 

Observe,  that  this  problem  is  an  application  of  (506,  Prob.  XI). 

(864) 


PRESENT     WORTH,  137 

EXAMPIiES     FOR     PRACTICE. 

624.  What  is  the  present  worth 

1.  Of  $800  at  6^,  due  in  6  mo.  ?    At  8^,  due  in  9  mo.  ? 

2.  Of  $360  at  7^,  due  in  2  yr,  ?    At  b%,  due  in  8  mo.  ? 

3.  Of  $490  at  %%,  due  in  42  da.  ?    At  1%,  due  in  128  da.  ? 

What  is  the  true  discount 

4.  Of  $580  at  t%,  due  in  90  da.  ?  At  8^^,  due  in  4  yr. 
17  da.? 

5.  Of  $260  at  %\%,  due  in  120  da.?  At  9^,  due  in  2  yr. 
25  da.? 

6.  Of  $860  at  7^,  due  in  93  da.  ?  At  12^,  due  in  3  yr. 
19  da.? 

7.  What  is  the  true  discount  at  ^%  on  a  debt  of  13200,  due 
in  2  yr.  5  mo.  and  24  da.  ? 

X  8.  Sold  my  farm  for  $3800  cash  and  a  mortgage  for  $6500 
running  for  3  years  without  interest.  The  use  of  money 
being  worth  1%  per  annum,  what  is  the  cash  value  of  the 
farm? 

9.  What  is  the  difference  between  the  interest  and  true  dis- 
count at  1%  of  $460,  due  8  months  hence  ? 

10.  A  man  is  offered  a  house  for  $4800  cash,  or  for  $5250 
payable  in  2  yr.  6  mo.  without  interest.  If  he  accepts  the 
former,  how  much  will  he  lose  when  money  is  worth  8^  ? 

11.  Which  is  more  profitable,  and  how  much,  to  buy 
wood  at  $4.50  a  cord  cash,  or  at  $4.66  payable  in  9  months 
without  interest,  money  being  worth  8;^? 

X 12.  A  merchant  buys  $2645.50  worth  of  goods  on  3  mo. 
credit,  but  is  offered  3^  discount  for  cash.  Which  is  the 
better  bargain,  and  how  much,  when  money  is  at  1%  per 
annum  ? 

13.  A  grain  merchant  sold  2400  bu.  of  wheat  for  $3600,  for 
which  he  took  a  note  at  4  mo.  without  interest.  What  was 
the  cash  price  per  bushel,  when  money  is  at  6^  ? 

(265) 


138  BUSIIfESS    ARITHMETIC, 


BANK    DISCOUNT. 

625.  ^anlz  Discount  is  the  interest  on  the  amount 
of  a  note  at  maturity,  computed  from  the  date  the  note  is 
discounted  to  the  date  of  maturity. 

1.  Observe  that  when  a  note  bears  no  interest,  its  amount  at  maturity 
is  its  face. 

2.  When  the  time  of  a  note  is  given  in  months,  calendar  months  are 
understood ;  if  the  month  in  which  the  note  falls  due  has  no  day  cor- 
responding to  the  date  of  the  note,  then  the  note  is  due  on  the  last  day 
of  that  month. 

3.  In  computing  bank  discount,  it  is  customary  to  reckon  the  time  in 
days. 

4.  When  a  note  becomes  due  on  Sunday  or  a  legal  holiday,  it  must  be 
paid  on  the  day  previous. 

636.  JDcujs  of  Grace  are  three  days  usually  allowed  by 
law  for  the  payment  of  a  note  after  the  expiration  of  the  time 
specified  in  it. 

627.  The  MaUirity  of  a  note  is  the  expiration  of  the 
time  for  which  it  is  made,  including  days  of  grace. 

628.  The  I^poceeds  or  Avails  of  a  note  is  the  sum 
left  after  deducting  the  discount. 

629.  A  Protest  is  a  declaration  in  writing  by  a  Notary 
Public,  giving  legal  notice  to  the  maker  and  endorsers  of  a 
note  of  its  non-payinent. 

630.  Prob.  XII.— To  find  the  bank  discount  and 
proceeds  of  a  note  for  any  g:iven  rate  and  time. 

Observe,  that  the  bank  discount  is  the  interest  on  the  amount  of  the 
note  at  maturity,  computed  from  the  time  the  note  is  discounted  to  the 
date  of  maturity  ;  and  the  proceeds  is  the  amount  of  the  note  at  maturity, 
minus  the  bank  discount.     Hence  the  following 

631.  EuLE. — J.  Fmcl  the  amount  of  the  note  at  ma- 
tioj^ity,  compute  the  interest  upon  this  sum  from  the 
date  of  discounting  the  note  to  the  date  of  Wjaturity ; 
the  result  is  the  hanh  discount. 

(266) 


BANK    DISCOUNT,  139 

II.  Subtract  the  hank  discount  from  the  amount  of 
the  note  at  maturity ;  the  remainder  is  the  proceeds. 

EXAl^IPIiES     FOR     PRACTICE. 

632.  What  are  the  bank  discount  and  proceeds  of  a  note 

1.  Of  ^280  for  3  mo.  15  da.  at  1%  ?    For  6  mo.  9  da.  at  S%  ? 

2.  Of  1790  for  154  da.  at  6%  ?    For  2  mo.  12  da.  at  7%  ? 

3.  Of  $1600  for  80  da.  at  7%  ?    For  140  da.  at  8^%  ? 

4.  What  is  the  difference  between  the  ba7ik  and  true  dis- 
count on  a  note  of  $1000  at  7%,  payable  in  90  days  ? 

5.  Valuing  my  horse  at  $212,  I  sold  him  and  took  a  note 
for  $235  payable  in  60  days,  which  I  discounted  at  the  bank. 
How  much  did  I  gain  on  the  transaction  ? 

^  6.  A  man  bought  130  acres  of  land  at  $16  per  acre.  He 
paid  for  the  land  by  discounting  a  note  at  the  bank  for 
$2140.37  for  90  da.  at  6^.    How  much  cash  has  he  left? 

Find  the  date  of  mattirityj  the  time,  and  th«  proceeds  of  the 
following  notes : 

7.  $480.90.  RocHESTEK,  N.  Y.,  Mar.  15,  1876. 
Seventy  days  after  date  I  promise  to  pay  to  the  order  of 

N.  L.  Sage,  four  hundred  eighty  t%  dollars  for  value  received. 
Discounted  Mar.  29.  DuNCAi^  MacVicar. 

8.  $590.  Potsdam,  N.  Y,  Map  18, 1876. 
Three  months  after  date  I  promise  to  pay  to  the  order  of 

Wm.  Flint,  five  hundred  ninety  dollars,  for  value  received. 
Discounted  June  2.  Peter  Henderson. 

9.  $1600.  Rome,  N.  Y.,  Jan.  19,  1876. 
Seven  months  after  date  we  jointly  and  severally  agree  to 

pay  James  Richards,  or  order,  one  thousand  six  hundred  dol- 
lars at  the  National  Bank,  Potsdam,  N.  Y.,  value  received. 
Discounted  May  23.  Robert  Button, 

James  Jackson. 
(267) 


140  BUSINESS    ARITHMETIC. 

633.  Prob.  XIII.— To  find  the  face  of  a  note  when 
the  proceeds,  time,  and  rate  are  given. 

Observe,  that  the  proceeds  is  the  face  of  the  note  minus  the  interest  on 
it  for  the  given  time  and  rate,  and  consequently  that  the  proceeds  must 
contain  $1  minus  the  interest  of  $1  for  the  given  time  and  rate  as  many 
times  as  there  are  dollars  in  the  face  of  the  note.     Hence  the  following 

634.  Rule. — Divide  the  given  proceeds  hy  the  pro- 
ceeds of  $1  for  the  given  time  and  rate ;  the  quotient  is 
the  face  of  the  note. 


fiXAMPLES     FOR     PRACTICE. 

635.  What  must  be  the  face  of  a  note  which  will  give 

1.  For  3  mo.  17  da.,  at  Q%,  I860  proceeds?  $290?  $530.80? 

2.  For  90  da.,  at  7^,  $450  proceeds  ?     $186.25  ?    $97.32  ? 

3.  For  73  da.,  at  $8^,  $234.60  proceeds?  $1800?    $506.94? 
v4.  "What  must  be  the  face  of  a  note  for  80  days,  at  t%,  on 

which  I  can  raise  at  a  bank  $472.86  ? 

6.  The  avails  of  a  note  for  50  days  when  discounted  at  a 
bank  were  $350.80 ;  what  was  the  face  of  the  note  ? 

6.  How  much  must  I  make  my  note  at  a  bank  for  40  da., 
at  7^,  to  pay  a  debt  of  $296.40  ? 

7.  A  merchant  paid  a  bill  of  goods  amounting  to  $2850  by 
discounting  three  notes  at  a  bank  at  '7%,  the  proceeds  of  each 
paying  one-third  of  the  bill ;  the  time  of  the  first  note  was 
60  days,  of  the  second  90  days,  and  of  the  third  154  days. 
What  was  i\iQ  face  of  each  note  ? 

8.  For  what  sum  must  I  draw  my  note  March  23, 1876,  for 
90  days,  so  that  when  discounted  at  7^  on  May  1  the  pro- 
ceeds may  be  $490  ? 

9.  Settled  a  bill  of  $2380  by  giving  my  note  for  $890  at 
30  days,  bearing  interest,  and  another  note  at  90  days,  which 
when  discounted  at  1%  will  settle  the  balance.  What  is  the 
face  of  the  latter  note  ? 

(268) 


'^^t  EXCHANGE 


636.  Exchange  is  a  method  of  paying  debts  or  other 
obligations  at  a  distance  without  transmitting  the  money. 

Thus,  a  merchant  in  Chicago  desiring  to  pay  a  debt  of  $1800  in  New 
York,  pays  a  bank  in  Chicago  $1800,  plus  a  small  per  cent  for  their  trouble, 
and  obtains  an  order  for  this  amount  on  a  bank  in  New  York,  which  he 
remits  to  his  creditor,  who  receives  the  money  from  the  New  York  bank. 

Exchange  between  places  in  the  same  country  is  called  Inland  or 
Domestic  Exchange,  and  between  different  countries  Foreign  Exchavge. 

637.  A  Draft  or  Bill  of  Exchange  is  a  written 
order  for  the  payment  of  money  at  a  specified  time,  drawn  in 
one  place  and  payable  in  another. 

1.  The  Drawer  of  a  bill  or  draft  is  the  person  who  signs  it;  the 
Drawee,  the  person  directed  to  pay  it ;  the  Payee,  the  person  to  whom 
the  money  is  directed  to  be  paid  ;  the  Jndorser,  the  person  who  transfers 
his  right  to  a  bill  or  draft  by  indorsing  it ;  and  the  Holder,  the  person 
who  has  legal  possession  of  it. 

2.  A  Sight  Draft  or  Bill  is  one  which  requires  payment  to  be  made 
when  presented  to  the  payor. 

3.  A  Time  Draft  or  BiU  is  one  which  requires  payment  to  be  made  at 
a  specified  time  after  date,  or  after  sight  or  being  presented  to  the  pay(yr. 

Three  days  of  graco  are  usually  allowed  on  bills  of  exchange. 

4.  The  Acceptance  of  a  bill  or  draft  is  the  agreement  of  the  party  on 
whom  it  is  drawn  to  pay  it  at  maturity.  This  is  indicated  by  writing 
the  word  "Accepted"  across  the  face  of  the  bill  and  signing  it. 

When  a  bill  is  protested  for  non-acceptance,  the  drawer  is  bound  to 
pay  it  immediately. 

5.  Foreign  bills  of  exchange  are  usually  drawn  in  duplicate  or  tripli- 
cate, and  sent  by  different  conveyances,  to  provide  against  miscarriage, 
each  copy  being  valid  until  the  bill  is  paid. 

(269) 


142  BUSIJVUSS     ARITHMETIC. 

638.  The  Par  of  Exchange  is  the  relative  value  of 
the  coins  of  two  countries. 

Thus,  the  par  of  exchange  between  the  United  States  and  England  is 
the  number  of  gold  dollars,  the  standard  unit  of  United  States  money, 
which  is  equal  to  a  pound  sterling,  the  standard  unit  of  English  money. 
Hence  $4.8665  =  £1  is  the  par  of  exchange. 


DOMESTIC   EXCHANGE. 

639.  Domestic  JExchange  is  a  method  of  paying 
debts  or  other  obligations  at  distant  places  in  the  same  coun- 
try, without  transmitting  the  money. 

Fortns  of  Sight  and  Time  Drafts, 

Third  National  Bank  op  Rochester,  ) 
$890.  Rochester,  N.  Y.,  May  4, 1876.  \ 

At  sight,  pay  to  the  order  of  Chas.  D.  McLean,  eight  hun- 
dred  ninety  dollars.  William  Roberts,  Cashier. 

To  the  Seventh  National  Bank,  ) 
New  York,  N.  Y.     \ 

This  is  the  usual  form  of  a  draft  drawn  by  one  bank  upon  another. 

12700.  '         Syracuse,  N.  Y.,  July  25,  1876. 

At  fifteen  days  sight,  pay  to  the  order  of  Taintor  Brothers 
&  Co.,  tioo  thousand  seven  hundred  dollars,  vahce  received,  and 
charge  the  same  to  the  account  of      ^    ^  Stewart  &  Co. 

To  the  Tenth  National  Bank,  ) 
New  York,  N.  Y.     \ 

1.  This  is  the  usual  form  of  a  draft  drawn  by  a  firm  or  individual 
upon  a  bank.    It  may  also  be  made  payable  at  a  given  time  after  date. 

2.  All  time  drafts  should  be  presented  for  acceptance  as  soon  as 
received.  When  the  cashier  writes  the  word  "Accepted,"  with  the 
date  of  acceptance  across  the  face,  and  signs  his  name,  the  bank  is 
responsible  for  the  payment  of  the  draft  when  due. 

(370) 


DOMESTIC    EXCHANGE.  143 


METHODS    OF    DOMESTIC    EXCHANGE. 

640.  First  Method. — Tlie  party  desiring  to  Jransmit 
mo7iey,  purchases  a  draft  for  the  amount  at  a  banic,  and  sends 
it  by  mail  to  its  destination. 

Observe  carefully  the  following: 

1.  Banks  can  sell  drafts  only  upon  others  in  which  they  have  deposits 
in  money  or  equivalent  security.  Hence  banks  throughout  the  country, 
in  order  to  give  them  this  facility,  have  such  deposits  at  centres  of  trade, 
such  as  New  York,  Boston,  Chicago,  etc. 

2.  A  Bank  Draft  will  usually  be  purchased  by  banks  in  any  part  of 
the  country,  in  case  the  person  offering  it  is  fully  identified  as  the  party 
to  whom  the  draft  is  payable.  Hence,  a  debt  or  other  liability  may 
be  discharged  at  any  place  by  a  draft  on  a  New  York  bank. 

3.  A  draft  may  be  made  payable  to  the  person  to  whom  it  is  sent,  or 
to  the  person  buying  it.  In  the  latter  case  the  person  buying  it  must 
write  on  the  back  "  Pay  to  the  order  of  "  (name  of  party  to  whom  it  is 
Bent),  and  sign  his  own  name. 

Second  Method. — /.  The  party  desiring  to  transmit  money, 
deposits  the  amotmt  in  a  bank  and  takes  a  certificate  of  deptosit, 
tvhich  he  sends  as  by  first  method.    Or, 

II.  If  he  has  a  deposit  already  in  a  bank,  subject  to  his  check 
or  order,  it  is  customary  to  send  his  check,  certified  to  be  good 
by  the  cashier  of  the  bank. 

This  method,  in  either  of  these  forms,  is  ordinarily  followed  in 
making  payments  at  a  distance  by  persons  in  New  York  and  other  large 
centres  of  trade.  Banks  in  such  places  have  no  deposits  in  cities  and 
villages  throughout  the  country,  and  hence  do  not  sell  drafts. 

Certificates  of  deposits  and  certified  checks  are. purchased  by  banks 
in  the  same  manner  as  bank  drafts. 

Third  Method. — The  party  desiring  to  transmit  money, 
obtains  a  Post  Office  order  for  the  amount  and  remits  it  as 
before. 

As  the  amount  that  can  be  included  in  one  Post  Ofiice  order  is  limited, 
this  method  is  restricted  in  its  application.  It  is  usually  employed  in 
remitting  small  sums  of  money. 

(371) 


144  BUSIN^BSS    ARITHME  TIC, 

Fourth  Method. — The  party  desiring  to  transmit  money, 
makes  a  draft  or  order  for  the  amount  upon  apartyoivinghim, 
at  the  place  tvhere  the  money  is  to  be  sent,  and  remits  this  as 
previously  directed. 

1.  By  this  method  one  person  is  snid  to  draw  upon  another.  Such 
drafts  should  be  presented  for  payment  as  soon  as  received,  and  if  not 
paid  or  accepted  should  be  protested  for  non-payment  immediately. 

2.  Large  business  firms  have  deposits  in  banks  at  business  centres,  and 
credit  with  other  business  firms  ;  hence,  their  drafts  are  used  by  them- 
selves and  others  the  same  as  bank  drafts. 

641.  The  Premium  or  Discount  on  a  draft  depends  chiefly 
on  the  condition  of  trade  between  the  place  where  it  is  pur- 
chased and  the  place  on  which  it  is  made. 

Thus,  for  example,  merchants  and  other  business  men  at  Buffalo  con- 
tract more  obligations  in  New  York,  for  which  they  pay  by  draft,  than 
New  York  business  men  contract  in  Buffalo ;  consequently,  banks  at 
Buffalo  must  actually  send  money  to  New  York  by  Express  or  other 
conveyance.  Hence,  for  the  expense  thus  incurred  and  other  trouble  in 
handling  the  money,  a  small  premium  is  charged  at  Buffalo  on  New 
York  drafts. 

EXAMPIiES     FOR     PRACTICE. 

642.  1.  What  is  the  cost  of  a  sight  draft  for  $2400,  at  |^ 
premium  ? 

Solution.— Cost  =  $2400  +  |%  of  $2400  =  $2416. 

2.  What  is  the  cost  of  a  draft  for  -^3200,  at  \%  premium  ? 
Solution.— Cost  =  $3300  +  i%  of  $3200  =  $3204. 

Find  the  cost  of  sight  draft 

3.  For  $834,  premium  2%.  6.  For  $1500,  discount  ^%. 

4.  For  $6300,  premium  ^%.  7.  For  $384.50,  discount  ^%. 

5.  For  $132.80,  premium  1%.       8.  For  $295.20,  discount  1|^. 
9.  The  cost  of  a  sight  draft  purchased  at  1^%  premium  is 

$493.29 ;  what  is  the  face  of  the  draft? 

Solution.— At  1|%  premium,  $1  of  the  face  of  the  draft  cost  $1,015. 
Hence  the  face  of  the  draft  is  as  many  dollars  as  $1,015  is  contained 
times  in  $493.29,  which  is  $486. 

(372) 


DOMESTIC     EXCHANGE.  145 

Find  the  face  of  a.  draft  which  cost 

10.  $575.41,  premium  2f^.  13.  $819.88,  discount  J^. 

11.  $731.70,  premium  1^%,  14.  $273,847,  discount  \%. 

12.  $483.20,  premium,  f^.  15.  $315.65,  discount  If ^. 

16.  What  is  the  cost  of  a  draft  for  $400,  payable  in  3  mo., 
premium  IJ^,  the  bank  allowing  interest  at  4^  until  the 
draft  is  paid  ? 

Solution.— A  sight  draft  for  $400,  at  \\%  premium,  costs  $406,  but 
the  bank  allows  interest  at  4%  on  the  face,  $400,  for  3  mo.,  which  is  $4. 
Hence  the  draft  will  cost 


Find  the  cost  of  drafts 

,  17.  For  $700,  premium  \%,  time  60  da.,  interest  at  Z%, 

18.  For  $1600,  premium  IJ^,  time  50  da.,  interest  at  4^. 

19.  For  $2460,  discount  f^,  time  90  da.,  interest  at  ^%. 

20.  For  $1800,  discount  1%,  time  30  da.,  interest  at  b%, 

21.  A  merchant  in  Albany  wishing  to  pay  a  debt  of  $498.48 
in  Chicago,  sends  a  draft  on  New  York,  exchange  on  New 
York  being  at  \%  premium  in  Chicago ;  what  did  he  pay  for 
the  draft  ? 

Solution. — Tlie  draft  cashed  in  Chicago  commands  a  premium  of  ^% 
on  its  face.  The  man  requires,  therefore,  to  purchase  a  draft  whose 
face  plus  ^%  of  it  equals  $498.48.  Hence,  according  to  (506 — 5),  the 
amount  paid,  or  face  of  the  draft,  is  $498.48  -f- 1.005  = 


22.  Exchange  being  at  98}  {1^%  discount),  what  is  the  cost 
of  a  draft,  time  4  mo. ,  interest  at  6%  ? 

23.  The  face  of  a  draft  which  was  purchased  at- 1^%  pre- 
mium is  $2500,  the  time  40  da.,  rate  of  interest  allowed  4^; 
what  was  its  cost  ? 

24.  My  agent  in  Detroit  sold  a  consignment  of  goods  for 
$8260,  commission  on  the  sale  2^%.  He  remitted  the  pro- 
ceeds by  draft  on  New  York,  at  a  premium  of  i%.  What  is 
the  amount  remitted  ? 

(273) 


146  BUSINESS    ARITHMETIC, 


FOREIGN    EXOHAISTGE. 

643.  Foreign  Exchange  is  a  method  of  paying  debts 
or  other  obligations  in  foreign  countries  without  transmitting 
the  money. 

OUerce,  that  foreign  exchange  is  based  upon  the  fact  that  different 
countries  exchange  products,  securities,  etc.,  with  each  other. 

Thus,  the  United  States  sells  wheat,  etc.,  to  England,  and  England  in 
return  sells  manufactured  goods,  etc.,  to  the  United  States.  Hence,  par- 
ties in  each  country  hecome  indebted  to  parties  in  the  other.  For  this 
reason,  a  merchant  in  the  United  States  can  pay  for  goods  purchased  in 
England  hy  buying  an  order  upon  a  firm  in  England  which  is  indebted  to 
a  firm  in  the  United  States. 

Form  of  a  Bill  or  Set  of  Exchange, 

£400.  New  York,  July  13,  1876. 

At  sight  of  this  First  of  Exchange  (second  and  third  of 
the  sa7ne  date  and  tenor  unpaid),  pay  to  the  order  of  E.  D. 
Blakeslee  Four  Hundred  Pounds  Sterling,  for  value 
received,  and  charge  the  same  to  the  account  of 

Williams,  Beown"  &  Co. 

To  Martin,  Williams  &  Co.,  London. 

The  person  purchasing  the  exchange  receives  three  bills,  which  he 
sends  by  different  mails  to  avoid  miscarriage.  When  one  has  been 
received  and  paid,  the  others  are  void. 

The  above  is  the  form  of  the  first  bill.  In  the  Second  Bill  the  word 
*•  First  "  is  used  instead  of  "  Second,"  and  the  parenthesis  reads,  "  First 
and  Tliird  of  the  same  date  and  tenor  unpaid."  A  similar  change  is 
made  in  the  Third  Bill. 

644.  Exchange  with  Europe  is  conducted  chiefly 
through  prominent  financial  centres,  as  London,  Paris,  Berlin, 
Antwerp,  Amsterdam,  etc. 

645.  Quotations  are  the  published  rates  at  which  bills 
of  exchange,  stocks,  bonds,  etc.,  are  bought  and  sold  in  the 
money  market  from  day  to  day. 

(274) 


FOREIGN    EXCHANGE, 


147 


These  quotations  give  the  market  gold  value  in  United  States  money 
of  one  or  more  units  of  the  foreign  coin. 

Thus,  quotations  on  London  give  the  value  of  £1  sterling  in  dollars ; 
on  Paris,  Antwerp,  and  Geneva,  the  value  of  $1  in  francs ;  on  Hamburg, 
Berlin,  Bremen,  and  Frankfort,  the  value  of  4  marks  in  cents;  on 
Amsterdam,  the  value  of  a  guilder  in  cents. 

646.  The  following  table  gives  the  par  of  exchange,  or  gold 
value  of  foreign  monetary  units,  as  published  by  the  Secretary 
of  the  Treasury,  January  1, 1876  : 

TABLE   OF   PAR  OF  EXCHAN"GE. 


COUNTBIES. 

MONETARY  UNIT. 

STANDARD. 

VATUE  IN 

U.  S.  MONET. 

Austria 

Florin 

Silver 

.45,  3 
.11),  3 
.96,  5 
.54,  5 
.96,  5 
$1.00 
.91,  8 
.91,  3 
.26,  8 
.91,  8 

4.97,  4 
.19,  3 

4.86,  6i 
.19.  3" 
.23,  8 
.09,  7 
.43,  6 
.19,  3 

1.00 
.99,  8 
.38,  5 
.26,  8 
.91,  8 

1.08 
.73,  4 

1.00 
.19,  3 
.26,  8 
.19,  3 
.^2,  9 
.11,  8 
,04,  3 
.91,  8 

Belt^ium 

Bolivia 

Franc 

Gold  and  silver. 
Gold  and  silver. 

Gold 

Gold 

Dollar             

Brazil 

Milreis  of  1000  reis. 

Peso 

Bogota 

Canada 

Dollar 

Gold : 

Central  America. 
Chili 

Dollar 

Silver 

Peso 

Gold 

Denmark 

Ecuador 

Crown 

Gold 

Dollar    .... 

Silver  

Effvpt 

Pound  of  100  piasters  . . 

Franc. .    

Pound  sterling 

Drachma 

Gold 

France 

Gold  and  silver. 
Gold 

Great  Britain. . . . 
Greece 

Gold  and  silver. 
Gold 

German  Empire  . 
Japan 

Mark 

Yen 

Gold 

India 

Rupee  of  10  annas 

Lira .    . 

Silver 

Italy 

Gold  and  silver. 

Gold 

Silver. 

Liberia 

Dollar 

Mexico 

Dollar 

Netherlands 

Norway 

Florin    

Gold  and  silver. 
Gold 

Crown 

Peru 

Dollar         

Silver        .    . 

Portugal 

Russia 

Sandwich  Islands 
Spain 

Milreis  of  1000  reis 

Rouble  of  100  copecks. . 

Dollar 

Peseta  of  100  centimes  . 

Crown 

Franc 

Gold 

Silver 

Gold 

Gold  and  silver. 
Gold 

Sweden 

Switzerland 

Tripoli 

Gold  and  silver. 
Silver 

Mahbub  of  20  piasters. . 
Piaster  of  16  caroubs. . . 
Piaster 

Tunis 

Silver 

Turkey 

Gold 

U.  S.  of  Colombia 

Peso..- 

Silver 

(375) 


148  BUSINESS     ARITHMETIC, 


METHODS    OF    DIRECT    EXCHANGE. 

647.  Direct  Exchange  is  a  method  of  making  pay- 
ments in  a  foreign  country  at  the  quoted  rate  of  exchange 
with  that  country. 

First  Method. — The  person  desiring  to  traiismit  the  money 
purchases  a  Set  of  Exchange  for  the  amoiuit  on  the  country 
to  tuhich  the  money  is  to  be  sent,  and  forwards  the  three  hills 
ly  different  7nails  or  routes  to  their  destination. 

Second  Method. — Hie  person  desiring  to  transmit  the 
money  instructs  his  creditor  in  the  foreig^i  country  to  draw 
upon  him,  that  is,  to  sell  a  set  of  exchange  upon  him, 
which  he  pays  in  his  ow7i  country  when  prese^ited, 

EXAMPLES    FOR     PRACTICES. 

648.  1.  What  is  the  cost  in  currency  of  a  bill  of  exchange 
on  Liverpool  for  £285  9s.  6d.,  exchange  being  quoted  at  $4.88, 
and  gold  at  1.12,  brokerage  i%  ? 

Solution.  —  1.  We  reduce 

£285  9s.  6d.  =  £285.475  the  9s.  6d.  to  a  decimal  of  £1. 

$4.88  X  285.475  =  11393.118        Hence  £285  9s.  6d.  =  £285.475. 

$1.1225  X  1393.118=  1563.77+  ^-  Since  £1  =  $4. 88,  £285.475 

must  be  equal  $4.88x285.475 
=  $1393.118,  the  gold  value  of  the  bill  without  brokerage. 

3.  Since  %\  gold  is  equal  $1.12  currency,  and  the  brokerage  is  ^%, 
the  cost  of  $1  gold  in  currency  is  $1.1225.  Hence  the  bill  cost  in  cur- 
rency $1.1225  X  1393.118  =  $1563.77  +  . 

What  is  tlie  cost  of  a  bill  on 

2.  London  for  £436  8s.  3d.,  sterling  at  4.84|,  brokerage  ^%  ? 

3.  Paris  for  4500  francs  at  .198,  brokerage  ^%  ? 

4.  Geneva,  Switzerland,  for  8690  francs  at  .189  ? 

5.  Antiverp  for  4000  francs  at  .175,  in  currency,  gold  at  1.09? 

6.  Amsterdam  for  8400  guilders  at  41  J,  brokerage  ^%  ? 

7.  Frankfort  for  2500  marks,  quoted  at  .974? 

(276) 


FOREIGN    EXCHANGE.  149 

8.  A  merchant  in  Boston  instructed  his  agent  at  Berlin  to 
draw  on  him  for  a  bill  of  goods  of  43000  marks,  exchange  at 
24J,  gold  being  at  1.08|,  brokerage  \% ;  what  did  the  mer- 
chant pay  in  currency  for  the  goods  ? 


METHODS    OF    INDIRECT   EXCHANGE. 

649.  Indirect  Exchange  is  a  method  of  making 
payments  in  a  foreign  country  by  taking  advantage  of  the  rate 
of  exchange  between  that  country  and  one  or  more  other 
countries. 

Observe  carefully  the  following : 

1.  The  advantage  of  indirect  over  direct  exchange  under  certain  finan- 
cial conditions  which  sometimes,  owing  to  various  causes,  exist  between 
different  countries,  may  be  shown  as  follows  : 

Suppose  exchange  in  New  York  to  be  at  par  on  London,  but  on  Paris 
at  17  cents  for  1  franc,  and  at  Paris  on  London  at  24  francs  for  £1.  With 
these  conditions,  a  bill  on  London  for  £100  will  cost  in  New  York 
$486.65  ;  but  a  bill  on  London  for  £100  will  cost  in  Paris  24  francs  x  100 
=  2400  francs,  and  a  bill  on  Paris  for  2400  francs  will  cost  in  New  York 
17  cents  x  2400  =  $408. 

Hence  £100  can  be  sent  from  New  York  to  London  by  direct  exchange 
for  $486.65,  and  by  indirect  exchange  or  through  Paris  for  $408,  giving 
an  advantage  of  $480.65  —  $408  =  $78.65  in  favor  of  the  latter  method. 

2.  The  process  of  computing  indirect  exchange  is  called  Arbitration 
of  Exchange.  When  there  is  only  one  intermediate  place,  it  is  called 
Simple  Arbitration  ;  when  there  are  two  or  more  intermediate  places,  it 
is  called  Compound  Arbitration. 

Either  of  the  following  methods  may  be  pursued: 
First  Mettlob.—  The  person  desiring  to  transmit  the  money 
may  huy  a  hill  of  exchange  for  the  amount  on  an  intermediate 
jdace^  which  he  sends  to  his  agent  at  that  place  ivith  instruc- 
tions to  huy  a  hill  with  the  proceeds  on  the  place  to  which  the 
money  is  to  he  sent,  atid  to  for  tear  d  it  to  the  proper  party. 
This  is  called  the  method  hy  remittance. 

(377) 


150  BUSINESS    ARITHMETIC, 

Second  Method. — The  person  desiring  tosend  the  money 
instructs  his  creditor  to  draw  for  the  amount  on  his  agent  at 
an  intermediate  place,  and  his  agent  to  draiu  upon  him  for  the 
same  amount. 

This  is  called  the  method  by  drawing. 

Third  Method. — TJie  person  desiring  to  send  the  money 
instructs  his  agent  at  an  intermediate  place  to  draio  upon  him 
for  the  amount,  and  buy  a  hill  on  the  place  to  which  the  money 
is  to  he  sent,  and  forward  it  to  the  proper  party. 

This  is  called  the  method  hy  drawing  and  remitting. 

These  methods  are  equally  applicable  when  the  exchange  is 
made  through  two  or  more  interniediate  places,  and  the  solu- 
tion of  examples  under  each  is  only  an  application  of  compound 
numbers  and  business.    Probs.  VIII,  IX,  X,  and  XI. 

BXAMPIiES     FOR     PRACTICE, 

650.  1.  Exchange  in  New  York  on  London  is  4.83,  and 
on  Paris  in  London  is  244- ;  what  is  the  cost  of  transmitting 
63994  francs  to  Paris  through  London  ? 

Solution.— 1.  We  find  tlie  cost  of  a  bill  of  exchange  in  London  for 
63994  francs.  Since  24^  francs  =  £1,  63994  -4-  24i  is  equal  the  number 
of  £  in  63994  francs,  which  is  £2612. 

2.  We  find  the  cost  of  a  bill  of  exchange  in  New  York  for  £2613. 
Since  £1  =  $4.83,  the  bill  must  cost  $4.83  x  2612  =  |12615.96. 

2.  A  merchant  in  !N"ew  York  wishes  to  pay  a  debt  in  Berlin 
of  7000  marks.  He  finds  he  can  buy  exchange  on  Berlin  at 
.25,  and  on  Paris  at  .18,  and  in  Paris  on  Berlin  at  1  murk  for 
1.15  francs.  Will  he  gain  or  lose  by  remitting  by  indirect 
excliange,  and  how  much  ? 

3.  "What  will  be  the  cost  to  remit  4800  guilders  from  New 
York  to  Amsterdam  through  Paris  and  London,  exchange 
being  quoted  as  follows:  at  New  York  on  Paris,  .]<S|;  at 
Paris  on  London,  24^  francs  to   a  £;   and  at  London  on 

(278) 


EQUATION    OF    PAYMENTS,  l5t 

Amsterdam,  12-J^  guilders  to  the  £.     How  much  more  would 
it  cost  by  direct  exchange  at  39^  cents  for  1  guilder  ? 

4.  An  American  residing  in  Berlin  wishing  to  obtain  $6000 
from  the  United  States,  directs  his  agent  in  Paris  to  draw  on 
Boston  and  remit  the  proceeds  by  draft  to  Berlin.  Exchange 
on  Boston  at  Paris  being  .18,  and  on  Berlin  at  Paris  1  mark 
for  1.2  francs,  the  agent's  commission  being  ^%  both  for 
drawing  and  remitting,  how  much  would  he  gain  by  drawing 
directly  on  the  United  States  at  24J  cents  per  mark  ? 

EQUATIOK"    OF    PAYMEJ^TS. 

651.  An  Account  is  a  written  statement  of  the  dehit 
and  credit  transactions  between  two  persons  with  their  dates. 

The  debit  or  left-hand  side  of  an  account  (marked  Dr)  shows  the 
sums  due  to  the  Creditor,  or  person  keeping  the  account ;  the  credit  or 
right-hand  side  (marked  Or.)  shows  the  sums  paid  by  the  Debtor,  or  per- 
son against  whom  the  account  is  made. 

653.  The  Balance  of  an  account  is  the  difference  be- 
tween the  sum  of  the  items  on  the  debit  and  credit  sides. 

653.  Equation  of  Payments  is  the  process  of  find- 
ing a  date  at  which  a  debtor  may  pay  a  creditor  in  one 
payment  several  sums  of  money  due  at  different  times,  with- 
out loss  of  interest  to  either  party. 

654.  The  Equated  Tfnie  is  the  date  at  which  several 
debts  may  be  equitably  discharged  by  one  payment. 

655.  The  Maturity/  of  any  obligation  is  the  date  at 
which  it  becomes  due  or  draws  interest. 

656.  The  Term  of  Credit  is  the  interval  of  time  from 
the  date  a  debt  is  contracted  until  its  maturity. 

657.  The  Average  Term  of  Credit  is  the  interval 
of  time  from  tlic  mat^lrity  of  the  first  item  in  an  account 
to  the  Equated  Tiine, 

(379) 


152  £  USIJVESS    ARITHMETIC, 


PBEP  ABATOR  J^     PROPOSITIONS. 

658,  The  method  of  settling  accounts  by  equation  of 
payments  depends  upon  the  following  propositions;  hence 
they  should  be  carefully  studied : 

Prop.  I. — Wlien,  hy  agreement,  no  interest  is  to  be  paid  on 
a  debt  from  a  specified  time,  if  any  part  of  the  amount  is  paid 
by  the  debtor,  he  is  entitled  to  interest  until  the  expiration  of 
the  specified  time. 

Thus,  A  owes  B  $100,  payable  in  12  months  without  interest,  which 
means  that  A  is  entitled  by  agreement  to  the  use  of  $100  of  B's  money 
for  13  months.  Hence,  if  he  pays  any  part  of  it  before  the  expiration  of 
the  12  months,  he  is  entitled  to  interest. 

Observe,  that  when  credit  is  given  without  charging  interest,  the 
profits  or  advantage  of  the  transaction  are  such  as  to  give  the  creditor 
an  equivalent  for  the  loss  of  the  interest  of  his  money. 

Prop.  II. — After  a  debt  is  due,  or  the  time  expires  for 
ivhich  by  agreement  no  interest  is  charged,  the  creditor  is 
ejititled  to  interest  on  the  amount  until  it  is  paid. 

Thus,  A  owes  B  $300,  due  in  10  days.  When  the  10  days  expire,  the 
$300  should  be  paid  by  A  to  B.  If  not  paid,  B  loses  the  use  of  the 
money,  and  is  hence  entitled  to  interest  until  it  is  paid. 

Prop.  III. —  When  a  term  of  credit  is  allowed  upon  any 
of  the  items  of  an  account,  the  date  at  which  such  items  are  due 
or  commence  to  draw  interest  is  found  by  adding  its  term  of 
credit  to  the  date  of  each  item. 

Thus,  goods  purchased  March  10  on  40  days'  credit  would  bo  due  or 
draw  interest  March  10  +  40  da.,  or  April  19. 

659.  Prob.  I. — To  settle  equitably  an  account  con- 
taining only  debit  items. 

R.  Bates  bought  merchandise  of  H.  P.  Emerson  as  follows : 
May  17,  1875,  on  3  months'  credit,  $265;  July  11,  on  25  days, 
$460 ;  Sept.  15,  on  65  days,  $650. 

(280) 


EQUATION     OF    PAYMENTS,  153 

Find  the  equated  time  and  the  amount  that  will  equitably 
settle  the  account  at  the  date  when  the  last  item  is  due,  1% 
interest  being  allowed  on  each  item  from  maturity. 

SOLUTION  BY  INTEREST  METHOD. 

1.  We  find  the  date  of  maturity  of  each  item  thus  : 

$265  on  3  mo.  is  due  May  17  +  3  mo.  =  Aug.  17 
§460  on  25  da.  is  due  July  11  +  25  da.  =  Aug.  5. 
$650  on  65  da.   is  due  Sept.  15  +  65  da.  =  Nov.  19. 

2.  As  the  items  of  the  debt  arc  due  at  these  dates,  it  is  evident 
that  when  they  all  remain  unpaid  until  the  latest  maturity,  H.  P.  Emer- 
son is  entitled  to  legal  interest 

On  $265  from  Aug.  17  to  Nov.  19  =    94  da. 
On  $460  from  Aug.    5  to  Nov.  19  =  106  da. 
The  $650  being  due  Nov.  19  bears  no  interest  before  this  date. 

3.  On  Nov.  19,  H.  P.  Emerson  is  entitled  to  receive  $1375,  the  sum  of 
the  items  of  the  debt  and  the  interest  on  $265  for  94  da.  plus  the  inter- 
est on  $460  for  106  da.  at  7%,  which  is  $14.12. 

Hence  the  account  may  be  equitably  settled  on  Nov.  19  by  R.  Bates 
paying  H.  P.  Emerson  $1375  +  $14.12  =  $1389.12. 

4.  Since  H.  P.  Emerson  is  entitled  to  receive  Nov.  19,  $1375  +  $14.12 
interest,  it  is  evident  that  if  he  is  paid  $1375  a  sufficient  time  before 
Nov.  19  to  yield  $14.12  interest  at  this  date,  the  debt  will  be  equitably 
settled.  But  $1375,  according  to  (596),  will  yield  $14.12  in  53  + a  frac- 
tion of  a  day. 

Hence  the  equated  time  of  settlement  is  Sept.  26,  which  is  53  days 
previous  to  Nov,  19,  the  assumed  date  of  settlement. 

SOLUTION  BY  PRODUCT  METHOD. 

1.  We  find  in  the  same  manner  as  in  the  interest  metJiod  the  dates  of 
maturity  and  the  number  of  days  each  item  bears  interest. 

2.  Assuming  Nov,  19,  the  latest  maturity,  as  the  date  of  settlement, 
it  is  evident  that  H.  P.  Emerson  should  be  paid  at  this  date  $1375,  the 
sum  of  the  items  of  the  account  and  the  interest  on  $265  for  94  days 
plus  the  interest  on  $460  for  106  days. 

3.  Since  the  interest  on  $265  for  94  days  at  any  given  rate  is  equal  to 
the  interest  on  $265  x  94,  or  $24910,  for  1  day  at  the  same  rate,  and  the 
interest  on  $460  for  100  days  is  equal  to  the  interest  on  $460  x  106,  or 

(381) 


154  BUSINESS    ARITHMETIC, 

$48100,  for  1  day,  the  interest  due  II.  P.  Emerson  Nov.  19  is  equal  the 
interest  of  §24910  +  $48760,  or  $73670,  for  1  day. 

4.  Since  the  interest  on  $73670  for  1  day  is  equal  to  tiie  interest  on 
$1375  for  as  many  days  as  $1375  is  contained  times  in  $73670,  which  is 
53|f  I,  it  is  evident  that  if  H.  P.  Emerson  receive  the  use  of  $1375  for 
53  days  previous  to  Nov.  19,  it  will  be  equal  to  the  interest  on  $73670 
for  1  day  paid  at  that  date.  Consequently,  11.  Bates  by  paying  $1375 
Sept.  26,  which  is  53  days  before  Nov.  19,  discharges  equitably  the 
indebtedness. 

Hence,  Sept.  26  is  the  equated  time,  and  from  Aug.  5  to  Sept.  26,  or 
53  days,  is  the  average  term  of  credit. 

Observe,  that  R.  Bates  may  discharge  equitably  the  indebtedness  in 
one  of  three  ways  : 

(1.)  By  paying  Nov.  19,  the  latest  maturity,  $1375,  tlie  sum  of  the  items 
of  the  account,  and  the  interest  o/ $73670/??*  1  day. 

In  this  case  the  payment  is  $1375  +  $14.12  interest  =  $1389.12. 

(2.)  By  paying  $1375,  the  sum  of  the  items  in  cash,  on  Sept.  26,  the 

EQUATED  TIME. 

(3.)  By  giving  his  note  for  $1375,  the  sum  of  the  items  of  the  account^ 
bearing  interest  from  Sept.  26,  the  equated  time. 

Observe  this  is  equivalent  to  paying  the  $1375  in  cash  Sept.  26, 

From  these  illustrations  we  obtain  the  following 

660.  EuLE. — /.  Find  the  date  of  maturity  of  each 
item. 

II.  Assujne  as  the  date  of  settlemejit  the  latest  ma- 
turity,  and  find  the  number  of  days  from  this  date  to 
the  maturity  of  each  item. 

In  case  the  indebtedness  is  discharged  at  the  assumed  date 
of  settlement : 

///.  Find  the  interest  on  each  item  from  its  maturity 
to  the  date  of  settlement.  The  sum  of  the  items  plus 
this  interest  is  the  amount  that  must  he  paid  the  creditor. 

In  case  the  equated  time  or  term  of  credit  is  to  be  found  and 
the  indebtedness  discharged  in  one  payment,  either  by  cash 
or  note : 

(282) 


EQUATION     OF    P  ATMENTS*  155 

IV.  Multiply  each  item  hy  the  number  of  days  from 
its  maturity  to  the  latest  maturity  in  the  account,  and 
divide  the  sum,  of  these  products  hy  the  sum  of  the 
items ;  the  quotient  is  the  number  of  days  which  must 
he  counted  bach  from  the  latest  maturity  to  give  the 
equated  time. 

V.  The  first  maturity  subtracted  from  the  equated 
time  gives  the  average  term  of  credit. 

EXAMPLES     FOR      PRACTICE. 

661.  1.  Henry  Ross  purchases  Jan.  1,  187G,  $1600  worth 
of  goods  from  James  Mann,  payable  as  follows:  April  1, 1876, 
$700;  June  1, 1876,  $400;  and  Dec.  1,  1876, 1500.  At  what 
date  can  he  equitably  settle  the  bill  in  one  payment  ? 

When  the  interval  between  the  maturity  of  each  item  and  the  date  of 
settlement  is  months,  as  in  this  example,  the  months  should  not  be 
reduced  to  days  ;  thus. 

Solution. — 1.  Assuming  that  no  payment  is  made  until  Dec.  1,  James 
Mann  is  entitled  to  interest 

On  ^700  for  8  mo.  =  $700  x  8  or  $5600  for  1  month. 
On  $400  for  6  mo.  =  $400  x  6  or  $2400  for  1  month. 
Hence  he  is  entitled  to  the  use  of  $8000  for  1  month. 

2.  $8000^  $160*0  =  5,  the  number  of  months  (659—4)  which  must 
be  counted  back  from  Dec.  1  to  find  the  equated  time,  which  is  July  1, 
Hence  the  bill  can  be  equitably  settled  in  one  payment  July  1,  1876. 

2.  A  man  purchased  a  farm  May  23,  1876,  for  $8600,  on 
which  he  paid  $2600,  and  was  to  pay  the  balance,  without 
interest,  as  follows:  Aug.  10,  1876,  $2500;  Jan.  4,  1877, 
$1500 ;  and  June  14,  1877,  $2000.  Afterwards  it  was  agreed 
that  the  whole  should  be  settled  in  one  payment.  At  what 
date  must  the  payment  be  made  ? 

3.  Bought  merchandise  as  follows:  Feb.  3,  1875,  $380; 
April  13,  $520  ;  May  18,  $260 ;  and  Aug.  12,  $350,  each  item 
on  interest  from  date.  What  must  be  the  date  of  a  note  for 
the  sum  of  the  items  bearing  interest  which  will  equitably 
settle  the  bill? 

(283) 


156 


BUSINESS    ARITHMETIC. 


Find  the  date  at  which  a  note  bearing  interest  can  be  given 
as  an  equitable  settlement  for  the  amount  of  each  of  the  fol- 
lowing bills,  each  item  being  on  interest  from  the  date  of 
purchase : 


4.  Purchased  as  follows 
July     9, 1876,  $380 
Sept.  13,     "     $270 
Nov.  34,     «     $840 
Dec.   29,     "     $260. 

6.  Purchased  as  follows: 
April  17, 1877,1185; 
June  24,  «  $250; 
Sept.  13,     «     $462. 


6.  Purchased  as  follows 
May    5,  1876,  $186 
Aug.  10,     "      $230 
Oct.  15,     «      $170 
Dec.  20,     "      $195. 

7.  Purchased  as  follows ; 
Aug.25, 1877,  $280; 
Oct.  10,     "     $193; 
Dec.  18,     "     $290. 


8.  Find  the  average  term  of  credit  on  goods  purchased  as 
follows:  Mar.  23,  ^$700,  on  95  da.  credit;  May  17,  $480,  on 
45  da.;  Aug.  25,  $690,  on  60  da. ;  and  Oct.  2,  $380  on  35  da. 

9.  Sold  A.  Williams  the  following  bills  of  goods :  July  1 0, 
$2300,  on  6  mo.  credit;  Aug.  15,  $900,  on  5  mo. ;  and  Oct.  13, 
$830,  on  7  mo.  What  must  be  the  date  of  note  for  the  three 
amounts,  bearing  interest  which  will  equitably  settle  the 
account  ? 

66^*  Prob.  II. — To  settle  equitably  an  account  con- 
taining both  debit  and  credit  items. 

Find  the  amount  equitably  due  at  the  latest  maturity  of 
either  the  debit  or  credit  side  of  the  following  account,  and 
the  equated  time  of  paying  the  balance : 


2>n 


B.  Whitney. 


Or, 


1876. 
Mar.  17 
May  10 
Aug.    7 

Tomdse.     .     .    . 
"      "       at  4  mo. 
**      «       at  2  mo. 

$400 
880 
540 

1876. 
Apr.  13 
June  15 

• 

By  cash  .... 
'♦   draft  at  30  da. 

$300 
450 

(284) 


EQUATION     OF    P  ATME  NTS.  I57 

Before  examining  the  following  solution,  study  carefvUy  the  three 
propositions  under  (658). 


SOLUTION  BY  PRODUCT  METHOD. 

Due.  Ami.  Days.   Products. 

Mar.   17.  400  x  204  =   81600 

Sept.  10.  380  X    27  =   10260 

Oct.      7.  540 


Paid.     Ami.   Days,   Products. 
Apr.  13.     200  X  177   =   35400 
July  15.     250  X     84   =   21000 


450  56400 

Total  debt,  $1320  $91860    Amt.  whose  Int.  for  1  da.  is  due  to  Creditor. 

Total  paid,       450  56400    Amt;  whose  Int  for  1  da.  is  due  to  Debtor. 

Balance,        $870  $35460    Bai.  whose  Int.  for  1  day  is  due  to  Creditor. 

Explanation.— Assuming  Oct.  7,  the  latest  maturity  on  either  side  of 
the  account,  as  the  date  of  settlement,  the  creditor  is  entitled  to  interest 
on  each  item  of  the  debit  side,  and  the  debtor  on  each  item  of  the  credit 
side  to  this  date  (658).  Hence,  we  find,  according  to  (059 — 3),  the 
amount  whose  interest  for  1  day  both  creditor  and  debtor  are  entitled  to 
Oct.  7. 

2.  The  creditor  being  entitled  to  the  most  interest,  we  subtract  the 
amount  whose  interest  for  1  day  tlie  debtor  is  entitled  to  from  the  cred- 
itor's amount,  leaving  $3540,  the  amount  whose  interest  for  1  day  the 
creditor  is  still  entitled  to  receive. 

3.  We  find  the  sum  of  the  debit  and  credit  items,  and  subtract  the 
latter  from  the  former,  leaving  $870  yet  unpaid.  This,  with  68  cents, 
the  interest  on  $3540,  is  the  amount  equitably  due  Oct.  7,  equal  f  870.68. 

4.  According  to  (658-4),  $3540  ^  $870  =  40«&,  the  number  of  days 
previous  to  Oct.  7  when  the  debt  can  be  discharged  by  paying  the  bal- 
ance, $870,  in  cash,  or  by  a  note  bearing  interest.  Hence  the  equated 
time  of  paying  the  balance  is  Aug.  27. 

The  following  points  regarding  the  foregoing  solution 
should  be  carefully  studied  : 

1.  In  the  given  example,  the  sum  of  the  debit  is  greater  than  the  sum 
of  the  credit  items ;  consequently  the  balance  on  the  account  is  due  to 
the  creditor.  But  the  balance  of  interest  being  also  due  him,  it  is  evi- 
dent that  to  settle  the  account  equitably  he  should  be  paid  the  $870 
before  the  assumed  date  of  settlement.  Hence  the  equated  time  of  pay- 
ing the  balance  must  be  before  Oct.  7. 

2.  Had  the  balance  of  interest  been  on  the  credit  side,  it  is  evident  the 
debtor  would  be  entitled  to  keep  the  balance  on  the  account  until  the 

(285) 


158  BUSINESS    ARITHMETIC. 

interest  upon  it  would  be  equal  the  interest  due  him.    Hence  the  equated 
time  of  paying  the  balance  would  be  after  Oct.  7. 

3.  Had  the  balance  of  the  account  been  on  the  credit  side,  the  creditor 
would  be  overpaid,  and  hence  the  balance  would  be  due  to  the  debtor. 

Now  in  case  the  balance  of  interest  is  also  on  the  credit  side  and  due 
to  the  debtor,  it  is  evident  that  to  settle  the  account  equitably  the  debtor 
should  be  paid  the  amount  of  the  balance  before  the  assumed  date  of 
settlement.    Hence  the  equated  time  would  be  before  Oct.  7, 

In  case  the  balance  of  interest  is  on  the  debtor  side,  it  is  evident  that 
while  the  creditor  has  been  overpaid  on  the  account,  he  is  entitled  to  a. 
balance  of  interest,  and  consequently  should  keep  the  amount  he  has 
been  overpaid  until  the  interest  upon  it  would  be  equal  to  the  interest 
due  him.     Hence  the  equated  time  would  be  after  Oct  7. 

4.  The  interest  method  given  (0«59)  can  be  used  to  advantage  in  find- 
ing the  equated  time  when  the  time  is  long  between  the  maturity  of  the 
items  and  the  assumed  date  of  settlement.  In  case  this  method  of  solu- 
tion is  adopted,  the  foregoing  conditions  are  equally  applicable 

From  these  illustrations  we  obtain  the  following 

663.  Rule. — I.  Find  the  maturity  of  each  item  on 
the  debit  and  credit  side  of  the  account. 

II.  Assume  as  the  date  of  settlei7ient  the  latest  ma- 
turity on  either  side  of  the  account,  and  find  the  numr- 
her  of  days  from  this  date  to  the  maturity  of  each,  on 
hoth  sides  of  the  account. 

III.  Multiply  each  debit  and  credit  item  hy  the  num* 
her  of  days  from  its  maturity  to  the  date  of  settlement, 
and  divide  the  balance  of  the  debit  and  credit  products 
hy  the  balance  of  the  debit  and  credit  items ;  the  quo- 
tient is  the  numher  of  days  the  ecfuated  tirne  is  from 
the  assumed  date  of  settlement. 

IT.  In  case  the  balance  of  items  and  balance  of 
interest  are  both  on  the  same  side  of  the  account,  siib- 
tract  this  number  of  days  from  the  assumed  date  of 
settlement,  but  add  it  in  case  theij  are  on  opposite  sides; 
the  result  is  the  equated  time. 

(286) 


EQUATION     OF    PAYMENTS. 


1^9 


EXAMPIiES     FOR     PRACTICE. 

664.     1.  Find  the  face  of  a  note  and  the  date  from  which 
it  must  bear  interest  to  settle  equitably  the  following  account : 

Dr.  James  Hai^d  m  acct.  tvith  P.  Anstead.  Cr. 


1876. 
Jan.     7 
May  11 
June    6 

To  mdse.  on  3  mo. 
*'      "       "  2  mo. 
"      "       "  5  mo. 

1430 
390 
570 

1876. 
Mar.  15 
May  17 
Aug.    9 

By  draft  at  90  da, 
"  cash  .... 
"  mdse.  on  30  da. 

1500 
280 
400 

2.  Equate  the  following  account,  and  find  the  cash  payment 
Dec.  7, 187G: 

Cr. 


Dr. 


William  Hekdersoi^". 


1876. 
Mar.  23 
May  16 
Aug.    7 

To  mdse. 

on  45  da. 
"  25  da. 
"  35  da. 

§^? 

1876. 
Apr.  16 
June  25 
July  13 

By  cash  .... 
"  mdse.  on  30  da. 
"  draft  at  60  da. 

$490 
650 
200 

3.  Find  the  equated  time  of  paying  the  balance  on  the  fol- 
lowing account : 

Dr.  Hugh  MacVicar.  Or. 


1876. 

1876. 

Jan.  13 

To  mdse.  on  60  da. 

$840 

Feb.  15 

By  note  at  60  da. 

$700 

Mar.  24 

"      "      •*  40  da. 

580 

Apr.  17 

"  cash    .... 

460 

June    7 

"      "      "     4mo. 

360 

June    9 

"  draft  at  30  da. 

1150 

July  14 

"      "      "  80da. 

730 

4.  I  purchased  of  Wm.  Rodgers,  March  10,  187G,  $930 
worth  of  goods ;  June  23,  $680;  and  paid,  April  3,  1870  cash, 
and  gave  a  note  May  24  on  30  days  for  %500.  What  must  be 
the  date  of  a  note  bearing  interest  that  will  equitably  settle 
the  balance  ? 

(287) 


160  BUSINESS    ARITHMETIC, 


REVIEW  AND  TEST  QUESTION'S. 

665.     1.  Define  Simple,  Compound,  and  Annual  Interest. 

2.  Illustrate  by  an  example  every  step  in  the  six  per  cent, 
method. 

3.  Show  that  1%%  may  be  used  as  conveniently  as  6^,  and 
write  a  rule  for  finding  the  interest  for  months  by  this  method. 

4.  Explain  the  method  of  finding  the  exact  interest  of  any 
sum  for  any  given  time.  Give  reasons  for  each  step  in  the 
process. 

5.  Show  by  an  example  the  difference  between  true  and 
hank  discount.     Give  reasons  for  your  answer. 

6.  Explain  the  method  of  finding  ih.Q  2)resent  worth. 

7.  Explain  how  the  face  of  a  note  is  found  when  the  pro- 
ceeds are  given.    Illustrate  each  step  in  the  process. 

8.  Define  Exchange,  and  state  the  difference  between  Do- 
mestic and  Foreign  Exchange. 

9.  State  the  difference  in  the  three  bills  in  a  Set  of 
Exchange. 

10.  What  is  meant  by  Par  of  Exchange  ? 

11.  State  the  various  methods  of  Domestic  Exchange,  and 
illustrate  each  by  an  example. 

12.  Illustrate  the  method  of  finding  the  cost  of  a  draft  when 
exchange  is  at  a  discount  and  brokerage  allowed.  Give 
reasons  for  each  step. 

13.  State  the  methods  of  Foreign  Exchange. 

14.  Illustrate  by  an  example  the  difference  between  Direct 
and  Indirect  exchange. 

15.  Define  Equation  of  Payments,  an  Account,  Equated 
Time,  and  Term  of  Credit. 

16.  Illustrate  the  Interest  Method  of  finding  the  Equated 
Time  when  there  are  but  debit  items. 

17.  State  when  and  why  you  count  forward  from  the 
assumed  date  of  settlement  to  find  the  equated  time. 

(288) 


[     RATIO 


PREPARATORY     PROPOSITIONS, 

666.  Two  numbers  are  compared  and  their  relation  detet' 
mined  hy  dividing  the  first  hy  the  second. 

For  example,  the  relation  of  $8  to  $4  is  determined  thus,  |8-f-$4  =  3. 
Observe,  the  quotient  3  indicates  that  for  every  (me  dollar  in  the  $4, 
there  are  two  dollars  in  the  $8. 

Be  particular  to  observe  the  following: 

1.  When  the  greater  of  two  numbers  is  compared  with  the 
less,  the  relation  of  the  numbers  is-  expressed  either  by  the 
relation  of  an  integer  or  of  a  mixed  number  to  the  unit  1,  that 
is,  by  an  improper  fraction  whose  denominator  is  1. 

Thus,  20  compared  with  4  gives  20  -i-  4  =  5  ;  that  is,  for  every  1  in 
the  4  there  are  5  in  the  30.  Hence  the  relation  of  30  to  4  is  that  of  the 
integer  5  to  the  unit  1,  expressed  fractionally  thus,  f . 

Again,  39  compared  with  4  gives  29  -^-  4  =  7^  ;  that  is,  for  every  1 
in  the  4  there  are  7^  in  29.  Hence,  the  relation  of  29  to  4  is  that  of  the 
mixed  number  7^  to  the  unit  1. 

2.  When  the  less  of  two  numbers  is  compared  with  the 
greater,  the  relation  is  expressed  by  a  proper  fraction. 

Thus,  6  compared  with  14  gives  6  -5-  14  =  ^^  =  |  (255) ;  that  is,  for 
every  3  in  the  6  there  is  a  7  in  the  14.  Hence,  the  relation  of  6  to  14  is 
that  of  3  to  7,  expressed  fractionally  thus,  f . 

Observe,  that  the  relation  in  this  case  may  be  expressed,  if  desired, 

as  that  of  the  unit  1  to  a  mixed  number.     Thus,  6  -J- 14  =  ^  =  ^ 

'^ 
(255) ;  that  is,  the  relation  of  6  to  14  is  that  of  the  unit  1  to  3^-, 

(289) 


163  BUSINESS    ARITUMETIC. 

EXAMPIiES     FOR     PRACTICES. 

667.  Find  orally  the  relation 

1.  Of  24  to  3.  5.  Of  113  to  9.  9.  Of  42  to  77. 

2.  Of  56  to  8.  6.  Of  25  to  100.  10.  Of  85  to  9. 

3.  Of  7G  to  4.  7.  Of  16  to  48.  11.  Of  10  to  1000. 

4.  Of  38  to  5.  8.  Of  13  to  90.  12.  Of  75  to  300. 

668.  Prop.  II. — No  numhers  can  it  compared  lut  those 
which  are  of  the  same  denominatio7i. 

Thus  we  can  compare  $8  witli  $3,  and  7  inches  with  3  inches,  but  we 
cannot  compare  $8  with  2  inches  (155—1). 

Observe  carefully  the  following : 

1.  Denominate  numbers  must  be  reduced  to  the  lowest 
denomination  named,  before  they  can  be  compared. 

For  example,  to  compare  1  yd.  2  ft.  with  1  ft.  3  in.,  both  numbers 
must  be  reduced  to  inches.  Thus,  1  yd.  2  ft. =60  in.,  1  ft.  3  in.  =  15  in., 
and  GO  in.  h-  15  in.  =  4  ;  hence,  1  yd.  2  ft.  are  4  times  1  ft.  3  in. 

2.  Fractions  must  be  reduced  to  ihQ  ^'amQ  fractional  denom- 
ination before  they  can  be  compared. 

For  example,  to  compare  3|  lb.  with  |  oz.  we  must  first  reduce 
the  3|  lb.  to  oz.,  then  reduce  both  members  to  the  same  fractional 
unit.  "  Thus,  (1)  U  lb.  =  56  oz. ;  (2)  56  oz.  =  if^  oz. ;  (3)  ^^  oz.  -^  ^  oz, 
=  i-f  0  =  45  (290) ;  hence,  the  relation  3^  lb.  to  |  oz.  is  that 
of  45  to  1. 

BXAMPLES     FOR     PRACTICE. 

669.  Find  orally  the  relation 

1.  Of  4  yd.  to  2  ft.  4.'  Of  |  to  |.  7.  Of  1^  pk.  to  3  bu. 

2.  Of  $2  to  25  ct.  5.  Of  I  to  1  J.  8.  Of  3  cd.  to  6  cd.  ft. 

3.  Of  2i  gal.  to  I  qt.  6.  Of  |  oz.  to  2  lb.  9.  Of  If  to  2J. 
Find  the  relation 

10.  Of  105  to  28.  13.  Of  $364  to  |4|. 

11.  Of  6 1  to  f  14.  Of  2  yd.  If  ft.  to|  in. 

12.  Of  9-J  bu.  to  ^  pk.  15.  Of  1  i  pt.  to  2|  gal. 

(290) 


EAT  10.  163 


DEFINITIONS. 

670.  A  Hatio  is  a  fraction  which  expresses  the  relatioTi 
which  the  first  of  two  numbers  of  the  same  denomination  has 
to  the  second. 

Thus  the  relation  of  $6  to  $15  is  expressed  by  f ;  that  is, 
16  is  I  of  $15,  or  for  every  $2  in  |6  there  are  $5  in  $15.  In 
like  manner  the  relation  of  $12  to  $10  is  expressed  by  |. 

671.  The  Special  Sign  of  Ratio  is  a  colon  (:). 

Thus  4 : 7  denotes  that  4  and  7  express  the  ratio  4^ ;  hence, 
4 : 7  and  f  are  two  ways  of  expressing  the  same  thing.  The 
fractional  form  being  the  more  convenient,  should  be  used 
in  preference  to  the  form  with  the  colon. 

672.  The  Terms  of  a  Hatio  are  the  numerator  and 
denominator  of  the  fraction  that  expresses  the  relation 
between  the  quantities  compared. 

The  first  term  or  numerator  is  called  the  Antecedent, 
the  second  term  or  denominator  is  called  the  Consequent, 

673.  A  Simple  Ratio  is  a  ratio  in  which  each  term  is 
a  single  integer.    Thus  9 : 3,  or  f ,  is  a  simple  ratio. 

674.  A  Compound  Ratio  is  a  ratio  whose  terms  are 
formed  by  multiplying  together  the  corresponding  terms  of 
two  or  more  simple  ratios. 

Thus,  multiplying  together  the  corresponding  terms  of  the  simple 

ratios  7  :  3  and  5  : 2,  we  have  the  compound  ratio  5x7:3x2  =  35:  6, 

,  »     ,.      ,,7      5      7x5      35 
or  expressed  fractionally  -  x  -  =  ~ — -  =  — . 
3       2       o  y.  4i        D 

Observe,  that  when  the  multiplication  of  the  corresponding  terms  is 

performed,  the  compound  ratio  is  reduced  to  a  simple  ratio. 

675.  The  Reciprocal  of  a  number  is  1  divided  by  that 
number.     Thus,  the  reciprocal  of  8  is  1  ~  8  =  ^. 

(291) 


164  'li^sT^^uss  arithmetic. 

676.  The  Heciprocal  of  a  Matio  is  1  divided  by  the 
ratio. 

Thus,  the  ratio  of  7  to  4  is  7  : 4  or  |,  and  its  reciprocal  is  1  -^  J  =  4, 
according  to  (291).  Hence  the  reciprocal  of  a  ratio  is  the  ratio  invert- 
ed, or  the  consequent  divided  by  the  antecedent. 

GHH.  A  Matio  is  in  its  Shnplest  Terms  when  the 
antecedent  and  consequent  are  prime  to  each  other. 

6*78.  The  MMuction  of  a  Matio  is  the  process  of 
changing  its  terms  without  changing  the  relation  they  express. 

Thus  |,  f ,  §,  each  express  the  same  relation. 

PKOBLEMS  ON  KATIO. 

679.  Since  every  ratio  is  either  a  proper  or  improper 
fraction,  the  principles  of  reduction  discussed  in  (235)  apply 
to  the  reduction  of  ratios.  The  wording  of  the  principles 
must  be  shghtly  modified  thus: 

Pein.  I. — The  terms  of  a  ratio  must  each  represent  units 
of  the  same  hind, 

Prin".  II. — Multiplying  both  terms  of  a  ratio  by  the  same 
number  does  not  change  the  value  of  the  ratio, 

Prin.  III. — Dividing  both  terms  of  a  ratio  by  the  same 
number  does  not  change  the  value  of  the  ratio. 

For  the  illustration  of  these  principles  refer  to  (235). 

680.  Prob.  I. — To  find  the  ratio  between  two  given 
numbers. 

Ex.  1.   Find  the  ratio  of  $56  to  $84. 

Solution. — Since,  according  to  (66G),  two  numbers  are  compared 
by  dividing  the  first  by  the  second,  we  divide  $56  by  $84,  giving 
$56  -^  $84  =  If ;   that  is,   $56   is   ||   of   $84.     Hence   the    ratio  of 

(292) 


RATIO,  •  165 

Ex.  2.  Find  the  ratio  of  1  yd.  2  ft.  to  1  ft.  3  in. 

Solution. — 1.  Since,  according  to  (668),  only  numbers  of  the 
same  denomination  can  be  compared,  we  reduce  botli  terms  to  inches, 
giving  GO  in.  and  15  in. 

2,  Dividing  60  in.  by  15  in.  we  have  60  in.  -f- 15  in.  =  4 ;  that  is, 
60  in.  is  4  times  15  in.    Hence  the  ratio  of  1  yd.  2  ft.  to  1  ft.  3  in.  is  f. 

EXAMPLES     FOR     PRACTICE. 

681.   Find  tlie  ratio 

1.  Of  $512  to  1256.  3.  Of  982  da.  to  2946  da. 

2.  Of  143  yd.  to  365  yd.  4.  Of  73  A.  to  365  A. 

5.  Of  £41  5s.  6d.  to  £2  3s.  6d. 

6.  Of  20  T.  6  cwt.  93  lb.  to  25  cwt.  43  lb.  5  oz. 

683.  Pbob.  II. — To  reduce  a  ratio  to  its  simplest 
terms. 

Reduce  the  ratio  ^  to  its  simplest  terms. 

Solution.— Since,  according  to  (679 — III),  the  ralue  of  the  ratio 
J/  is  not  changed  by  dividing?  both  terms  by  tlie  same  number,  we 
divide  the  antecedent  15  and  the  consequent  9  by  3,  their  greatest 

common  divisor,  giving  -^  \  o  =  6-    ^^^  having  divided  15  and  9  by 

their  greatest  common  divisor,  the  quotients  5  and  3  must  be  prime  to 
each  other.    Hence  (677)  f  are  the  simplest  terms  of  the  ratio  -^/. 


E^XAMPIiES     FOR     PRACTICES. 

683.  Reduce  to  its  simplest  terms 

1.  The  ratio  6  :  9.  4.  The  ratio  fff. 

2.  The  ratio  21  :  56.  5.  The  ratio  65  :  85. 

3.  The  ratio  ^.  6.  The  ratio  195  :  39. 

Express  in  its  simplest  terms  the  ratio  (see  668) 

7.  Of  96  T.  to  56  T.  9.  Of  8s.  9d.  to  £1. 

8.  Of  f  ft.  to  2  yd.  10.  Of  3  pk.  5  qt.  to  1  bu.  2  pk. 

(293) 


166  •      BUSINESS    ARITHMETIC. 

684.  Prob.  III. — To  find  a  number  that  has  a  given 
ratio  to  a  given  number. 

How  many  dollars  are  |  of  $72  ? 

Solution. — The  fraction  |  denotes  the  ratio  of  the  required  number 
to  $73  ;  namely,  for  every  $8  in  $72  there  are  $5  in  the  required 
number.  Consequently  we  divide  the  $72  by  $8,  and  multiply  $5 
by  the  quotient.  Hence,  first  step,  $72  -^  $8  =  9  ;  second  step  $5x9 
=  $45,  the  required  number. 

Observe,  that  this  problem  is  the  same  as  Prob.  VIII,  501,  and 
Prob.  II,  274.  Compare  this  solution  with  the  solution  in  each  of 
these  problems. 

SXAMPIiES     FOR     PRACTICE. 

685.  Solve  and  explain  each  of  the  following  examples, 
regarding  the  fraction  in  every  case  as  a  ratio. 

1.  How  many  days  are  -^  of  360  days  ? 

2.  A  man  owning  a  farm  of  243  acres,  sold  -^  of  it ;  how 
many  acres  did  he  sell  ? 

3.  James  has  $'i'96  and  John  has  f  as  much  ;  how  much 
has  John  ? 

4.  A  man's  capital  is  $4500,  and  he  gains  -^  of  his  capital ; 
how  much  does  he  gain  ? 

5.  Mr.  Jones  has  a  quantity  of  flour  worth  $3140 ;  part  of 
it  being  damaged  he  sells  the  whole  for  -J-f  of  its  value;  how 
much  does  he  receive  for  it  ? 

686.  Prob.  IV. — To  find  a  number  to  which  a  given 
number  has  a  given  ratio. 

$42  are  J  of  how  many  dollars  ? 

Solution.— The  fraction  |  denotes  the  ratio  of  $42  to  the  required 
number ;  namely,  for  every  $7  in  $42  there  are  $4  in  the  required 
number.  Consequently  we  divide  the  $42  by  $7  and  multiply  $4  by 
the  quotient.  Hence,  first  step,  $42  -^  $7  =  G  ;  second  step,  $4x0  =  $24, 
the  required  number. 

Observe,  that  this  problem  is  the  same  as  Prob.  IX,  502.  Compare 
the  solutions  and  notice  the  points  of  diflference. 

(394) 


RATIOi  •  IW 


EXAMPLES     FOR     PRACTICE. 

687.  Solve  and  explain  each  of  the  following  examples, 
regarding  the  fraction  in  every  case  as  a  ratio. 

1.  96  acres  are  f|  of  how  many  acres  ? 

2.  I  received  $75,  which  is  f  of  my  wages ;  how  much  is 
still  due  ? 

3.  James  attended  school  117  days,  or  -f^  of  the  term  ; 
how  many  days  in  the  term  ? 

4.  Sold  my  house  for  $2150,  which  was  |^  of  what  I  paid 
for  it ;  how  much  did  I  lose  ? 

5.  Henry  reviewed  249  lines  of  Latin,  or  |  of  the  term's 
work ;  how  many  hnes  did  he  read  during  the  term  ? 

6.  48  cd.  3  cd.  ft.  of  wood  is  y\  of  what  I  bought ;  how 
much  did  I  buy  ? 

7.  Mr.  Smith's  expenses  are  J  of  his  income.  He  spends 
$1500  per  year ;  what  is  his  income  ? 

8.  4  gal.  3  qt.  1  pt.  are  ^  of  how  many  gallons  ? 

9.  A  merchant  sells  a  piece  of  cloth  at  a  profit  of  12.50, 
which  is  -^  of  what  it  cost  him;  how  much  did  he 
pay  for  it  ? 

688.  Prob.  V. — To  find  a  number  to  which  a  given 
number  has  the  same  ratio  that  two  other  given  num- 
bers have  to  each  other. 

To  how  many  dollars  have  $18  the  same  ratio  that  6  yd. 
have  to  15  yd.? 

Solution.— 1.  We  find  by  (680—1)  the  ratio  of  6  yd.  to  15  yd., 
which  is  j^j  =  I,  according  to  (077). 

2.  Since  |  denotes  the  ratio  of  the  $18  to  the  required  number,  the 
$18  must  be  the  antecedent ;  hence  we  have,  according  to  (086),  first 
step,  $18  -4-  $2  =  9  ;  second  step,  $5x9  =  $45,  the  required  number. 

Observe,  that  in  this  problem  we  have  the  antecedent  of  a  ratio  given 
to  find  the  consequent.  In  the  foUov/ing  we  have  the  consequent  given 
to  find  the  antecedent. 

(295) 


168  BUSINESS    ARITH3IETIC, 

689.  Prob.  VI. — To  find  a  number  that  has  the  same 
ratio  to  a  given  number  that  two  other  given  numbers 
have  to  each  other. 

How  many  acres  have  the  same  ratio  to  12  acres  that  $56 
have  to  $84  ? 

Solution.— 1.  We  find  by  (680— I)  the  ratio  of  $56  to  $84,  which 
is  tf  =  I,  according  to  (677). 

3.  Since  f  denotes  the  ratio  of  the  required  number  to  12  acres, 
the  13  acres  must  be  the  consequent  ;  hence  we  liave,  according  to 
(684),  first  step,  13  acr.  -r-  3  acr.  =  4 ;  second  step,  3  acr.  x  4  =  8  acres, 
the  required  number. 

EXAMPLES     FOR     PRACTICE. 

690.  The  following  are  applications  of  Prob.  V  and  VL 

1.  If  12  bu.  of  wheat  cost  $15,  what  will  42  bu.  cost? 

Ecgarding  the  solution  of  examples  of  this  kind,  observe 
that  the  price  or  rate  per  unit  is  assumed  to  be  the  same  for 
each  of  the  quantities  given. 

Thus,  since  the  12  bu.  cost  $15.  the  price  per  bushel  or  unit  is  $1.25, 
and  the  example  asks  for  the  cost  of  43  bu.  at  this  price  per  bushel. 
Consequently  whatever  part  the  13  bu.  are  of  43  bu.,  the  $15.  the  cost  of 
13  bu.,  must  be  the  same  part  of  the  cost  of  43  bu.  Hence  we 'find  the 
ratio  of  13  bu.  to  43  bu.  and  solve  the  example  by  Prob.  V. 

2.  What  will  16  cords  of  wood  cost,  if  2  cords  cost  $9  ? 

3.  If  a  man  earn  $18  in  2  weeks,  how  much  will  he  earn 
in  52  weeks  ? 

4.  If  24  bu.  of  wheat  cost  $18,  what  will  36  bu.  cost  ? 

5.  If  24  cords  of  wood  cost  $60,  what  will  40  cords  cost  ? 

6.  Bought  170  pounds  of  butter  for  $51;  what  would  680 
pounds  cost,  at  the  same  price  ? 

7.  Two  numbers  are  to  each  other  as  10  to  15,  and  the  less 
number  is  329 ;  what  is  the  greater? 

8.  At  the  rate  of  16  yards  for  $7,  how  many  yards  of  cloth 
can  be  bought  for  $100  ? 

(296) 


1      .      '^y^^      ..,      1 


'-M^[[PROPORTIONl]|^i^ 


DEFINITIONS. 

691.  A  Proportion  is  an  equality  of  ratios,  the  terms 
of  the  ratios  being  expressed. 

Thus  the  ratio  f  is  equal  to  the  ratio  ^f  ;  hence  f  =  ^f  is  a  propor- 
tion, and  is  read,  The  ratio  of  3  to  5  is  equal  to  the  ratio  of  12  to  20, 
or  3  is  to  5  as  12  is  to  20. 

692.  The  equality  of  two  ratios  constituting  a  proportion 
is  indicated  either  by  a  double  colon  (: :)  or  by  the  sign  (=). 

Thus,  I  =  ^y,  or  3  :  4  =  9  :  12,  or  3  :  4  : :  9  :  12. 

693.  A  Simple  JProportion  is  an  expression  of  the 
equality  of  two  simple  ratios. 

Thus,  t\  =  f  f »  or  8  :  12  : :  3  :  48,  or  8  :  12  =  3  :  8  is  a  shnple  pro- 
portion.   Hence  a  simple  proportion  contains  four  terms. 

694.  A  Compound  Proportion  is  an  expression 
of  the  equality  of  a  compound  (674)  and  a  simple  ratio  (673). 

Thus,  «  i  K  [  : :  48  :  60,  or  f  X  I  =  f§,  is  a  compound  proportion.  It 
is  read,  The  ratio  of  2  into  6  is  to  3  into  5  as  48  is  to  60. 

695.  A  Proportional  is  a  number  used  as  a  term  in 
a  proportion. 

Thus  in  the  simple  proportion  2:5  : :  6  :  15  the  numbers  2,  5,  6,  and 
15  are  its  terms ;  hence,  each  one  of  these  numbers  is  called  a  propor- 
tional, and  the  four  numbers  together  are  called  proportionals. 

(397) 


170  BUSINESS    ARITHMETIC, 

696.  A  Mean  Proportional  is  a  number  that  is  the 
Consequent  of  one  and  the  Antecedent  of  the  other  of  the  two 
ratios  forming  a  proportion. 

Thus  in  the  proportion  4  :  8  : :  8  :  16,  the  number  8  is  the  consequent 
of  the  first  ratio  and  the  antecedent  of  the  second ;  hence  is  a  mean 
proportioned. 

697.  The  Antecedents  of  a  proportion  are  the  first  and 
third  terms,  and  the  Consequents  are  the  second  and 
fourth  terms. 

698.  The  Extremes  of  a  proportion  are  its  first  and 
fourth  terms,  and  the  Means  are  its  second  and  third 
terms, 

SIMPLE    PROPORTIOK. 

PRJEPARATOUT     STEPS, 

699.  The  following  preparatory  steps  should  be  perfectly- 
mastered  before  applying  proportion  in  the  solution  of 
problems.  The  solution  of  each  example  under  Step  I 
should  be  given  in  full,  as  shown  in  (688  and  689),  aud 
Step  II  and  III  should  be  illustrated  by  the  pupil,  .in  the 
manner  shown,  by  a  number  of  examples. 

700.  Step  I. — Find  by  Prob.  V  and  VI,  in  ratio,  the 
missing  term  in  the  following  proportions : 

The  required  term  is  represented  by  the  letter  x. 

1.  6  :  42  : :  5  :  ic.  4.    5  bu.  2  pk.  :  3  pk.  : :  a:  :  4  bn. 

2.  24  :  60  : :  a;  :  15.         5.    2  yd.  :  8  in.  : :  a;  :  3  ft.  4  in. 

3.  84  :  a;  : :  21  :  68.         6.    a;  :  £3  2s.  : :  49  T.  :  18  cwt. 

Step  II. — Show  that  the  product  of  the  extremes  of  apro^ 
portion  is  equal  to  the  product  of  the  means. 

Thus  the  proportion  3  :  3  : :  G  :  9  expressed  fractionally  gives  f  =  f . 

(298) 


SIMPLE    PROPORTION.  171 

Now  if  both  terms  of  this  equality  be  multiplied  by  3  and  by  9,  the 
consequents  of  the  given  ratios,  the  equality  is  not  changed  ;  hence, 

2jr9  >^  ^  62^x_9^     Cancelling  (186)  the  factor  3  in  the  left-hand 

3  9 

term  and  9  in  the  right-hand  term  we  have  3x9  =  6x3.  But  2  and  9 
are  the  extremes  of  the  proportion  and  6  and  3  are  the  means ;  hence  the 
truth  of  the  proposition. 

Step  III. — Sliow  thai,  since  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means,  any  term  of  a  proportion 
can  he  found  ivhen  the  other  three  are  known. 

Thus  in  the  proportion  3  :  a;  : :  9  :  15  we  have  known  the  two 
extremes  3  and  15  and  the  mean  9.  But  by  Step  II,  3  x  15,  or  45,  is 
equal  to  9  times  the  required  mean ;  hence  45  -f-  9  =  5,  the  required 
mean.  In  the  same  manner  any  one  of  the  terms  may  be  found  ;  hence 
the  truth  of  the  proposition. 

Find  by  this  method  the  missing  term  in  the  following : 

$13  :  a;  : :  5  yd.  :  3  yd. 
128  bii.  :  3  pk.  : :  a;  :  $1.25. 
64  cwt.  :  X  ::  $120  :  $15. 


Solution  by  Simple  Proportion, 

701.  The  quantities  considered  in  problems  that  occur  in 
practical  business  are  so  related  that  when  certain  conditions 
are  assumed  as  invariable,  they  form  ratios  that  must  be  equal 
to  each  other,  and  hence  can  be  stated  as  a  proportion  thus. 

If  4  yd.  of  cloth  cost  $10,  what  will  18  yd.  cost  ? 

Observe,  that  in  this  example  the  price  per  yard  is  assumed 
to  be  invariable,  that  is,  the  price  is  the  same  in  both  cases ; 
consequently  whatever  part  the  4  yd.  are  of  the  18  yd.,  the  $10 
are  the  same  part  of  the  cost  of  the  18  yd.,  hence  the  ratio  of 
the  4  yd.  to  the  18  yd.  is  equal  the  ratio  of  the  $10  to  the 
required  cost,  giving  the  proportion  4  yd.  :  18  yd.  : :  $10  :  %x, 

(299) 


1. 

14  :  3  : :  a;  :  12. 

4 

2. 

tc  :  24  : :  7  :  8. 

5, 

3. 

27  :  iz;  : :  9  :  5. 

6. 

172 


BUSINESS    ARITH3IETIC, 


EXAItIPr.ES     FOR     PRACTICE. 

702.  Examine  carefully  the  following  proportions  and 
state  what  must  be  considered  in  each  case  as  invariable,  and 
why,  in  order  that  the  proportion  may  be  correct. 


1. 


8. 


The  number 

'  The  number 

("The  cost  1 

f 

The  cost 

of  units 

is  to 

of  units 

-«J    ^^*^^    \isto 

in  the 

bought  in 

bought  in 

first 

second 

one  case     - 

another  case- 

I     case     J 

case. 

The      ■ 

r       The 

The       ■ 

r       The 

Principal 

.        Principal 

.. 

interest  in 

interest  in 

in  one 

1  in  another 

the  first 

1  the  second 

case      J         I       case      J 

case       J         I       case. 

The  number  ' 

The  number 

'     The 

'     The 

of  men 

of  men 

number 

number 

that  can  do 

:   . 

that  can  do 

, , 

of  days 

, 

of  days 

a  piece  of 

the  same 

the 

the 

work  in 

work  in 

second 

first 

one  case 

.  another 

case  J 

.    work    . 

.  work. 

Wliy  is  the  second  ratio  of  this  proportion  made  the  ratio  of  the 
number  of  days  the  second  work  to  the  number  of  days  the  first  work  ? 
Illustrate  this  arrangement  of  the  terms  of  the  ratio  by  other  examples. 

In  solving  examples  by  simple  proportion,  the  following 
course  should  be  pursued : 

7.  Represent  the  required  term  hj  x,  and  maTce  it  the  last 
extreme  or  consequent  of  the  seco?id  ratio  in  the  jjroportion. 

II.  Find  the  term  in  the  example  that  is  of  the  same  denom- 
ination as  the  required  term,  and  maJce  it  the  second  mean  or 
the  antecedent  of  the  second  ratio  of  the  proportion. 

III.  Determine,  by  inspecting  carefully  the  conditio7is  given 
in  the  example,  whether  x,  the  required  term  of  the  ratio  now 
expiressed,  must  be  greater  or  less  than  the  given  term^ 

(300) 


SIMPLE    PROPORTION.  173 

IV.  If  X,  the  required  term  of  the  ratio  expressed,  must  he 
greater  than  the  given  term,  make  the  greater  of  the  remaining 
terms  in  the  example  the  consequent  of  the  first  ratio  of  the 
proportion  ;  if  less,  make  it  the  antecedent. 

V.  When  the  proportion  is  stated^  find  the  required  term 
either  as  shown  in  (688)  or  in  (689). 

Observe,  that  in  either  way  of  finding  the  required  term,  any  factor 
that  is  common  to  the  given  extreme  and  either  of  the  given  means 
should  be  cancelled,  as  shown  in  (186). 

4.  How  maoy  bushels  of  wheat  would-  be  required  to  make 
39  barrels  of  flour,  if  15  bushels  will  make  3  barrels? 

5.  If  77  pounds  of  sugar  cost  $8.25,  what  will  84  pounds  cost? 

6.  I  raised  245  bushels  of  corn  on  7  acres  of  land;  how 
many  bushels  grow  on  2  acres  ? 

7.  If  6  men  put  up  73  feet  of  fence  in  3  days,  how  many 
feet  will  they  put  up  in  33  days  ?• 

8.  What  will  168  pounds  of  salt  cost,  if  3J^  pounds  cost 
37i  cents  ? 

9.  If  25  cwt.  of  iron  cost  $84.50,  what  will  24f  cwt.  cost  ? 

10.  Paid  $2225  for  18  cows,  and  sold  them  for  $2675;  what 
should  I  gain  on  120  cows  at  the  same  rate  ? 

11.  If  5  lb.  10  oz.  of  tea  cost  $5.25  ;  w^hat  will  7  lb.  8  oz.  cost  ? 

12.  If  a  piece  of  cloth  containing  18  yards  is  worth  $10.80, 
what  are  4  yards  of  it  worth  ? 

13.  My  horse  can  travel  2  mi.  107  rd.  in  20  minutes ;  how 
far  can  lie  travel  in  2  hr.  20  min. 

14.  If  18  gal.  3  qt.  1  pt.  of  water  leaks  out  of  a  cistern  in 
4  hours,  how  much  will  leak  out  in  36  hours  ? 

15.  Bought  28  yards  of  cloth  for  $20;  what  price  per  yard 
would  give  me  a  gain  of  $7.50  on  the  whole  ? 

16.  If  I  lend  a  man  $69.60  for  8|-  months,  how  long  should 
he  lend  me  $17.40  to  counterbalance  it? 

17.  My  annual  income  on  U.  S.  6^'s  is  $337.50  when  gold 
is  at  112^ ;  what  would  it  be  if  gold  were  at  125  ? 

(301) 


in 


BUSINESS    ARITHMETIC, 


COMPOUND    PROPORTION. 

PRE1*A.RATOJtY     STEPS. 

703,  Step  I. — A  compound  ratio  is  reduced  to  a  simple 
one  hy  m^ultiplying  the  antecedents  together  for  an  antecedent 
and  the  consequents  for  a  consequent  (6*74). 

i  6  •  7  ) 
Thus  the  compound  ratio  )  j^  1  o  f  ^^  reduced  to  a  simple  ratio  by 

multiplying  the  antecedents  6  and  4  together,  and  the  consequents 
7  and  3.  Expressing  the  ratios  fractionally  we  "have  f  x  f  =  |f  =  f 
(68^). 

Observe,  that  any  factor  that  is  common  to  any  antecedent  and 
consequent  may  he  cancelled  before  the  terms  are  multiplied. 

Reduce  the  following  compound  ratios  to  simple  ratios  in 
their  simplest  terms, 

35 

L16J 

Step  II. — A  compound  proportion  is  reduced  to  a  simple 
proportion  by  reducing  the  compound  ratio  to  a  simple  ratio. 

Thus,  in  the  compound  proportion  ]  «  .\£  [  : :  24  :  18,  the  com- 
pound ratio  f  X  f  is  equal  the  simple  ratio  f  ;  substituting  this  in  tlm 
proportion  for  the  compound  ratio  we  have  the  simple  proportion 
4  :  3  : :  24  :  18. 

Observe,  that  when  a  compound  proportion  is  reduced  to  a  simpl 
proportion,    the    missing    term    is    found    according    to    (688),    o. 
(689) 


1.    \ 


r  ^  • 

25-^ 

15  : 

18 

28  : 

50 

.   3  : 

7, 

rl6  : 

9- 

27: 

15 

28 
^      8 

Find  the  missing  term  in  the  following  : 

( 24  :  15  )  (  48  :  20 

1.    •<    7  :  16  [    : :  40  :  a;.  2.   •]        3 
(  25  :  21  )  (6 

(302) 


\- 


:  28  :  a;. 


COMPOUND     PROPORTION.  175 

Solution  by  Compound  Proportion, 

704.  The  following  preparatory  propositions  should  be 
carefully  studied  and  the  course  indicated  observed  in  solving 
l)roblems  involving  compound  proportion. 

Prop.  I. — There  are  one  or  more  conditions  in  every  example 
involving  proportion,  wliicli  must  he  regarded  as  invariable 
in  order  that  a  solution  may  he  given,  thus 

If  9  horses  can  subsist  on  50  bu.  of  oats  for  20  days,  how  long  can 
6  horses  subsist  on  70  bu. 

In  this  example  there  are  two  conditions  that  must  be  considered  as 
invariable  in  order  to  give  a  solution  : 

1.  The  fact  that  each  horse  subsists  on  the  same  quantity  of  oats 
each  day. 

3.  The  fact  that  each  bushel  of  oats  contains  the  same  amount 
of  food. 

Prop.  II. — To  solve  a  proUem  involving  a  compound  pro- 
portion, the  effect  of  each  ratio,  ivhich  forms  the  compound 
ratio,  on  the  required  term  must  he  considered  separately, 

thus : 

If  5  men  can  build  40  yards  of  a  fence  in  12  days,  how 
many  yards  can  8  men  build  in  9  days. 

1.  We  observe  that  the  invariahle  co7iditions  in  this  example 
are 

(1.)  That  each  man  in  both  cases  does  the  same  a/mount  of  work 
in  the  same  time. 

(2.)  That  the  same  amount  of  work  is  required  in  each  case  to  bvHd 
one  yard  of  the  fence. 

2.  We  determine  by  examining  the  problem  how  the  re- 
quired term  is  affected  by  the  relations  of  the  given  term,  thus : 

(1.)  We  observe  that  the  5  men  in  12  days  can  build  40  yards. 
Now  since  each  man  can  build  the  same  extent  of  the  fence  in  one" 
day,  it  is  evident  that  if  the  8  men  work  12  days  the  same  as  the 
5  men,  the  40  yards  built  by  the  5  men  in  13  days  must  have  the 

(303) 


176  BUSTNUSS    ARITHMETIC, 

same  ratio  to  the  number  of  yards  that  can  be  built  by  the  8  men 
in  12  days  as  5  men  have  to  8  men ;  hence  the  proportion 

5  men  :  8  men  : :  40  yards  :  x  yards. 

This  proportion  will  give  the  number  of  yards  the  8  men  can 
build  in  12  days. 

(2.)  We  now  observe  that  the  8  men  work  only  9  days ;  and  since 
they  can  do  the  same  amount  of  work  each  day,  the  work  done  in 
12  days  must  have  the  same  ratio  to  the  work  they  can  do  in  9  days 
that  12  days  have  to  9  days.     Hence  we  have  the  compound  proportion 

5  men  :  8  men )         .^        ,  •, 

13  days:  9  days  [    =  ^  40  yards  :  0=  sards. 

We  find  from  this  proportion,   according  to  (703 — II),  that  the 

8  men  can  build  48  yards  of  fence  in  9  days. 

EXAMPLES     FOR     PRACTICJB. 

705.  1.  If  it  cost  188  to  hire  12  horses  for  5  days,  what 
will  it  cost  to  hire  10  horses  for  18  days  ? 

2.  If  12  men  can  saw  45  cords  of  wood  in  3  days,  working 

9  hours  a  day,  how  much  can  4  men  saw  in  18  days,  working 
12  hours  a  day  ? 

3.  If  28  horses  consume  240  bushels  of  corn  in  112  days, 
how  many  bushels  will  12  horses  consume  in  196  days  ?. 

4.  When  the  charge  for  candying  20  centals  of  grain 
50  miles  is  $4.50,  what  is  the  charge  for  carrying  40  centals 
100  miles? 

5.  The  average  cost  of  keeping  25  soldiers  1  year  is  $3000 ; 
what  would  it  cost  to  keep  139  soldiers  7  years  ? 

6.  If  1  pound  of  thread  makes  3  yards  of  linen,  IJ  yard 
wide,  how  many  pounds  would  make  45  yards  of  linen, 
1  yard  wide  ? 

7.  G4  men  dig  a  ditch  72  feet  long,  4  feet  wide,  and  2  feet 
deep,  in  8  days ;.  how  long  a  ditch,  2 J  feet  wide  and  IJ  feet 
deep,  can  96  men  dig  in  60  days  ? 

8.  If  it  requires  8400  yd.  of  cloth  1^  yd.  wide  to  clothe  3500 
soldiers,  how  many  yards  |  wide  will  clothe  6720  ? 

(304) 


-*>e  H^^^SJ^r^g)^  j^ 


PARTNERSHIP 


DEFINITIONS. 

706.  A  I^artnevshii)  is  an  association  of  two  or  more 
persons  for  the  transaction  of  business. 

The  persons  associated  are  called  partners,  and  the  Association  is 
called  a  Company,  Firm,  or  House. 

707.  The  Capital  is  the  money  or  other  property  invest- 
ed in  the  business. 

The  Capital  is  also  called  the  Investment  or  Jdnt-stock  of  the 
Company. 

708.  The  Assets  or  Effects  of  a  Company  are  the 
property  of  all  kinds  belonging  to  it,  together  with  all  the 
amounts  due  to  it. 

709.  The  lAahilities  of  a  company  are  its  debts. 


PBEPARATOBT     P  B  O  P  O  8  1  T I O  If  8  . 

710.  Prop.  I. — 77ie  profits  and  the  losses  of  a  com' 
pany  are  divided  among  the  partners,  according  to  the 
value  of  each  marCs  investment  at  the  time  the  division 
is  made. 

Observe  carefully  the  following  regarding  this  proposition: 

Since  the  use  of  money  or  property  is  itself  value,  it  is  evident  that 
the  value  of  an  investment  at  any  time  after  it  is  made,  depends  first  upon 

(305) 


178  BUSITTESS    ARITHMETIC. 

the  amount  invested,  second  on  the  length  of  the  time  the  investment 
has  been  made,  and  third  the  rate  of  interest. 

Thus  the  value  of  an  investment  of  $500  at  the  time  it  is  made  is  just 
$500  ;  but  at  the  end  of  9  years,  reckoning  its  use  to  be  worth  7  %  per 
annum,  its  mine  will  be  |500  +  $315  =  $815. 

Prop.  II. — Tlie  value  of  any  investment  made  for  a  given 
number  of  intervals  of  time,  can  he  represented  by  another 
investment  made  for  one  interval  of  time. 

Thus,  for  example,  the  value  of  an  investment  of  $40  for  5  months  at 
any  given  rate  of  interest  is  the  same  as  the  value  of  5  times  $40,  or  $200, 
for  one  month. 


SXAMPLBS     FOR     practice:. 

711.  Find  the  value  at  simple  interest 

1.  Of  $800  invested  4  years  at  6^  per  annum. 

2.  Of  1350  invested  2  yr.  3  mo.  at  7;^  per  annum. 

3.  Of  $2860  invested  19  months  at  S%  per  annum. 

Solve  the  following  by  applying  (710 — II). 

4.  An  investment  of  1200  for  6  months  is  equal  in  value  to 
what  investment  for  4  months? 

5.  A  man  invests  1600  for  9  months,  $700  for  3  months,  and 
$300  for  7  months,  each  at  tlie  same  rate  of  interest.  What 
sum  can  he  invest  for  4  months  at  the  given  rate  of  interest, 
to  be  equal  in  value  to  the  three  investments  ? 

IZZZrSTRATION     OF     PROCESS . 

713.  Prob.  I. — To  apportion  grains  or  losses  when 
each  partner's  capital  is  invested  the  same  length  of 
time. 

Observe,  that  when  each  partner's  capital  is  used  for  the  same  length 
of  time,  it  is  evident  that  his  share  of  the  gain  or  loss  must  be  the  same 
fraction  of  the  whole  gain  or  loss  that  his  capital  is  of  the  whole  capital. 
Hence,  examples  under  this  problem  may  be  solved— 

(306) 


PARTNERSHIP,  179 

1.  By  Proportion  thus : 

The  whole  \       {Each  man's  \         (  Whole  \       (       Each 
capital     y  :  ]      capital      y  ::  Xgain  or>  :  \man'sgain 
invested    )       (     invested    )         (     loss    )       \      or  loss, 

II.  By  Percentage  thus  : 

Find  loliat  per  cent  (504)  the  whole  gain  or  loss  is  of  the 
whole  capital  invested,  and  take  the  same  per  cent  of  each  man's 
investment  as  his  share  of  the  gain  or  loss, 

III.  By  Fractions  thus : 

Find  what  fractional  part  each  man's  investment  is  of  the 
imhole  capital  invested,  and  take  the  same  fractional  part  of  the 
gain  or  loss  as  each  man's  share  of  the  gain  or  loss, 

EXAMPLES     FOR     PRACTICE. 

713.  1.  Three  men,  A,  B,  and  C,  form  a  company ;  A 
puts  in  $G000;  B  $4000;  and  C  $5600;  they  gain  $4320; 
what  is  each  man's  share  ? 

2.  A  man  failing  in  business  owes  A  $9600,  B  $7000,  and 
C  $5400,  and  his  available  property  amounts  to  $5460 ;  what  is 
each  man's  share  of  the  property  ? 

3.  Three  men  agree  to  liquidate  a  church  debt  of  $7890, 
each  paying  in  proportion  to  his  property ;  A's  property  is 
valued  at  $6470,  B'sat  $3780,  and  C's  at  $7890;  what  portion 
of  the  debt  does  each  man  pay  ? 

4.  A  building  worth  $28500  is  insured  in  the  ^tna  for 
$3200,  in  the  Home  for  $4200,  and  in  the  Mutual  for  $0500; 
it  having  been  partially  destroyed,  the  damage  is  set  at  $10500; 
what  should  each  company  pay  ? 

5.  The  sum  of  $2600  is  to  be  divided  among  four  school 
districts  in  proportion  to  the  number  of  scholars  in  each ; 
in  the  first  there  are  108,  in  the  second  84,  in  the  third  72, 
in  the  fourth  48 ;  what  part  should  each  receive  ? 

(307) 


180  BUSINESS    ARITHMETIC. 

714.  Prob.  II. — To  apportion  gains  or  losses  when 
each  partner's  capital  is  invested  dift'erent  lengths  of 
time. 

Observe  carefully  the  following : 

1.  According  to  (710 — II)  we  can  find  for  eacli  partner  an  amount 
whose  value  invested  one  interval  of  time  is  equal  to  the  value  of  his 
capital  for  the  given  intervals  of  time. 

2.  Having  found  this  we  can,  by  adding  these  amounts,  find  an  amount 
whose  value  invested  one  interval  of  time  is  equal  to  the  total  value  of 
the  whole  capital  invested. 

When  this  is  done  it  is  evident  that  each  man's  share  of  the  gain 
or  loss  must  be  the  same  fraction  of  the  whole  gain  or  loss  that  the 
mlue  of  his  investment  is  of  the  total  value  of  the  whole  capital 
Invested.  Hence  the  problem  from  this  point  can  be  solved  by  either 
of  the  three  methods  given  un(Jer  Prob.  I  (712). 

EXAMPLES    FOR     PRACTICES. 

715.  1.  A  and  B  engage  in  business ;  A  puts  in  $1120  for 
5  months  and  B  $480  for  8  months ;  they  gain  1354  ;  what  is 
each  man's  share  of  the  gain  ? 

2.  Three  men  hire  a  pasture  for  1136.50 ;  A  puts  in  16  cows 
for  8  weeks,  B  puts  in  6  cows  for  12  weeks,  and  C  the  same 
number  for  8  weeks;  what  should  each  man  pay  ? 

3.  The  joint  capital  of  a  company  was  17800,  which  was 
doubled  at  the  end  of  the  year.  A  put  in  ^  for  9  mo.,  B  ^  for 
8  mo.,  and  C  the  remainder  for  1  year.  What  is  each  one's 
stock  at  the  end  of  the  year  ? 

4.  Jan.  1,  1875,  three  persons  began  business.  A  put  in 
$1200,  B  put  in  $500  and  May  1  $800  more,  C  put  in  $700 
and  July  1  $400  more ;  at  the  end  of  the  year  the  profits  were 
$875 ;  how  shall  it  be  divided? 

5.  A  and  B  formed  a  partnership  Jan.  1,  1876.  A  put  in 
$6000  and  at  the  end  of  3  mo.  $900  more,  and  at  the  end  of 
10  mo.  drew  out  $300 ;  B  put  in  $9000  and  8  mo.  after  $1500 
more,  and  drew  out  $500  Dec.  1 ;  at  the  end  of  the  year  the 
aet  profits  were  $8900.    Find  the  share  of  each. 

(308) 


alligation]  f^^s^^ 


ALLIGATION  MEDIAL. 

716.  Alligation  Medial  is  the  process  of  finding  the 
mean  or  average  price  or  quality  of  a  mixture  composed  of 
several  ingredients  of  different  prices  or  qualities. 

BXAMPIiSS     FOR     PRACTICE. 

717.  1.  A  grocer  mixed  7  lb.  of  coffee  worth  30  ct.  a 
pound  witli  4  lb.  @  25  ct.  and  10  lb.  @  32 ;  in  order  that  he 
may  neither  gain  or  lose,  at  what  price  must  he  sell  the 
mixture  ? 

7  lb.  @  30  ct.  ~  $2.10  Solution.— 1.  Since  the  value  of 

4  lb.  @  25  ct.  =     1.00  G^^  kind  of  coffee  is  not  changed 

10  lb!  @  32  ct  =    3^20         ^^  °''^'''^'  ^^  ^^^  *^'^  ""^^^^  ""^  *^'® 
■ entire  mixture  by  finding  the  value 

KL  lb.  =  ^o.oO  of  each  kind  at  the  given  price,  and 

A/.  Of)  _i_  0-1   OA  pj-  taking  the  sum  of  these  values  as 

shown  in  illustration. 

2.  Having  found  that  the  21  lb.  of  coifee  are  worth  at  the  given 

prices  |6.30,  it  is  evident  that  to  realize  this  amount  from  the  sale  of 

the  21  lb.  at  a  uniform  price  per  pound,  he  must  get  for  each  pound  ^^^ 

of  $0.30  ;  hence,  .$6.30  -t-  21  =  30  cents,  the  selling  price  of  the  mixture. 

2.  A  wine  merchant  mixes  2  gallons  of  wine  worth  11.20  a 
gallon  with  4  gallons  Avorth  $1.40  a  gallon,  4  gallons  worth 
$.90  and  8  gallons  worth  8.80  a  gallon  ;  what  is  the  mixture 
worth  per  gallon  ? 

(309) 


182  BUSINESS    ARITHMETIC. 

3.  A  grocer  mixes  48  lb.  of  sugar  at  17  ct.  a  pound  with 
58  lb.  at  13  ct.  and  94  lb.  at  11  ct.;  what  is  a  pound  of  the 
mixture  worth  ? 

4.  A  goldsmith  melts  together  6  ounces  of  gold  22  carats 
fine,  30  ounces  20  carats  fine,  and  12  ounces  14  carats  fine ; 
how  many  carats  fine  is  the  mixture  ? 

5.  A  merchant  purchased  60  gallons  of  molasses  at  30  ct. 
per  gallon  and  40  gallons  at  25  cents,  which  he  mixed  with 
8  gallons  of  water.  He  sold  the  entire  mixture  so  as  to  gain 
20  per  cent  on  the  original  cost ;  what  was  his  selling  price 
per  gallon  ? 

ALLIGATION  ALTERNATE. 

718.  Alligation  Alternate  is  the  process  of  finding 
the  proportional  quantities  of  ingredients  of  different  prices 
or  quahties  that  must  be  used  to  form  any  required  mixture, 
when  the  price  or  quality  of  the  mixture  is  given. 


PREPARA.TORT     PROPOSITIONS, 

719.  Prop.  I. — In  forming  any  mixture,  it  is  assumed  that 
the  value  of  the  entire  mixture  must  he  equal  to  the  aggregate 
value  of  its  ingredients  at  their  given  prices. 

Thus,  if  10  pounds  of  tea  at  45  ct.  and  5  pounds  at  60  ct.  be  mixed, 
the  value  of  the  mixture  must  be  the  value  of  the  10  pounds  plus  the 
value  of  the  5  pounds  at  the  given  prices,  which  is  equal  $4.50  +  $3.00 
=  $7.50.    Hence  there  is  neither  gain  or  loss  in  fonning  a  mixture. 

Prop.  II. — The  price  of  a  mixture  must  he  less  than  the 
highest  and  greater  than  the  lowest  price  of  any  ingredient 
used  informing  the  mixture. 

Thus,  if  sugar  at  10  ct.  and  at  15  ct,  per  pound  be  mixed,  it  is  evident 
the  price  of  the  mixture  must  be  less  than  15  cents  and  greater  than 
10  cents ;  that  is,  it  must  be  some  price  between  10  and  15  cents. 

(310) 


ALLIGATION,  \%% 

ILLVSTBATION     OF     rRQCESS. 

720.  If  tea  at  56  ct.,  60  ct,  75  ct.,  and  90  ct.  per  pound  be 
mixed  and  sold  at  66  ct.  per  pound,  how  much  of  each  kind 
of  tea  can  be  put  in  the  mixture  ? 

First  Step  in  Solution, 

We  find  the  gain  or  loss  on  one  unit  of  each  ingredient  thus  : 

.    .     (66  ct.  —  56  ct.  =  10  ct.  gain. 
^    '    I  66  ct.  —  60  ct.  =    6  ct.  gain. 

75  ct.  —  66  ct.  =:    9  ct.  loss. 

90  ct.  —  QQ  ct.  =  24  ct.  loss. 


(^.)    { 


Second  Step  in  Solution, 

We  now  take  an  ingredient  on  which  there  is  a  gain,  and  one  on  which 
there  is  a  loss,  and  ascertain  how  much  of  each  must  be  put  in  the  mix- 
ture to  make  the  gain  and  loss  equal ;  thus  : 

Producing  Gain.  Gained  and  Lost.  Pboditcing  Loss. 

(1.)    9  lb.  at  10  ct.  per  lb.  gain.    =    90  ct.   =    10  lb.  at     9  ct.  per  lb.  loss. 
(2.)   4  lb.  at  24  ct.  per  lb.  gain.    =    24  Ct.    =      1  lb.  at  24  ct.  per  lb.  loss. 

Hence  the  mixture  must  contain  9  lb.  at  56  cts.  per  pound,  10  lb.  at 
75  ct.  per  pound,  4  lb.  at  60  ct.  per  pound,  and  1  lb.  at  9  ct.  per  pound. 

731.  Observe  carefully  the  following: 

1.  The  gain  and  loss  on  any  two  ingredients  may  be 
balanced  by  assuming  any  amount  as  the  sum  gained  and  lost. 

Thus,  instead  of  taking  90  cents,  as  in  (1)  in  the  above  solution,  as  the 
amount  gained  and  lost,  we  might  take  360  cents ;  and  dividing  360  cents 
by  10  cents  would  give  36,  the  number  of  pounds  of  56  ct.  tea  that 
would  gain  this  sum.  Again,  dividing  360  cents  by  9  cents  would  give 
40,  the  number  of  pounds  of  75  ct.  tea  that  would  lose  this  sum. 

2.  To  obtain  integral  proportional  parts  the  amount  assumed 
must  be  a  multiple  of  the  gain  and  loss  on  one  unit  of  the 
ingredients  balanced,  and  to  obtain  the  least  integral  propor- 
tional parts  it  must  be  the  least  common  multiple. 

(311) 


184  BUSINJSSS    ARITHMETIC. 

3.  When  a  number  of  ingredients  are  given  on  which  there 
is  a  gain  and  also  on  which  there  is  a  loss,  they  may  be 
balanced  with  each  other  in  several  ways ;  hence  a  series  of 
different  mixtures  may  be  formed  as  follows: 

Taking  the  foregoing  example  we  have 

A  Second  Mixture  thus: 

Prodtjcing  Gain.  Gained  and  Lost.  Producing  Loss. 

(1.)  24  lb.  at  10  Ct.  per  lb.  gain.    —    240  ct.  =    10  lb.  at  24  Ct.  per  lb.  loss. 
(2.)     9  lb.  at     C  ct.  per  lb.  gain.    :zz      54  ct.  =       G  lb.  at    9  ct.  per  lb.  loss. 

Hence  the  mixture  is  composed  of  24  lb.  @  56  ct.,  9  lb.  @,  60  ct., 
10  lb.  @  90  ct,  and  6  lb.  @  75  ct. 

A  Tliird  Mixture  thus: 

Pboducing  Gain.             Gained  and  Lost.  Pboducing  Loss. 

(1.)     9  lb.  at  10  ct.  per  lb.  gain.    ==      90  Ct.  =  10  lb.  at    9  ct.  per  lb.  loss. 

(2.)  24  lb.  at  10  ct.  per  lb.  gain.    =    240  ct.  =  10  lb.  at  24  ct.  per  lb.  loss. 

(3.)     9  lb.  at    Get.  per  lb.  gain.    =:      54  ct.  =  6  lb.  at    9  Ct.  per  lb.  loss. 

Observe,  that  in  (1)  and  (2)  we  have  balanced  the  loss  on  the  75  ct,  and 
90  ct.  tea  by  the  gain  on  the  56  ct.  tea  ;  hence  we  have  9  lb.  +  24  lb. ,  or 
33  lb.  of  the  56  ct.  tea  in  the  mixture. 

Observe,  also,  that  in  (3)  we  have  balanced  the  ga.in  on  the  60  ct.  tea 
by  a  loss  on  the  75  ct.  tea  ;  hence  we  have  10  lb.  +  6  lb,,  or  16  lb.  of  the 
75  ct.  tea  in  the  mixture. 

Hence  the  mixture  is  composed  of  33  lb.  @  56  ct.,  9  lb.  @  60  ct., 
16  lb.  @  75  ct.,  and  10  lb.  @  90  ct. 

4.  Mixtures  may  be  formed  as  follows : 

/.  Take  any  pair  of  ingredients,  one  giving  a  gain  and  the 
other  a  loss,  and  find  the  gain  and  loss  on  one  unit  of  each, 

11.  Assume  the  least  common  multiple  of  the  gain  and  loss 
on  one  unit  as  the  amount  gained  and  lost,  hy  putting  the  tiuo 
ingredients  in  the  mixture. 

IIL  Divide  the  amount  thus  assumed  hy  the  gain  and  then 
by  the  loss  on  one  unit ;  the  results  will  be  respectively  the 

(313) 


ALLIG  ATION,  185 

number  of  units  of  each  i7igredient  that  must  be  in  the  mixture 
that  the  gain  and  loss  may  balance  each  other. 

I V.  Proceed  in  the  same  manner  with  other  ingredients  j  the 
results  ivill  be  the  proportional  parts, 

EXAMPLES     FOR     PRACTICK. 

732.  1.  How  much  sugar  at  10,  9,  7,  and  5  ct.  will  pro- 
duce a  mixture  worth  8  cents  a  pound  ? 

2.  A  man  wishes  to  mix  sufficient  water  with  molasses 
worth  40  cents  a  gallon  to  make  the  mixture  worth  24  cents 
a  gallon ;  what  amount  must  he  take  of  each  ? 

3.  A  jeweller  has  gold  16,  18,  22,  and  24  carats  fine;  how 
much  of  each  must  he  use  to  form  gold  20  carats  fine  ? 

4.  A  merchant  desires  to  mix  flour  worth  $6,  $7|-,  and  $10 
a  barrel  so  as  to  sell  the  mixture  at  $9 ;  what  proportion  of 
each  kind  can  he  use  ? 

6.  A  farmer  has  wheat  worth  40,  55,  80,  and  90  cents  a 
bushel ;  how  many  bushels  of  each  must  be  mixed  with  290 
@40  ct.  to  form  a  mixture  worth  70  cents  a  bushel? 

Examples  like  this  where  the  quantity  of  one  or  more 
ingredients  is  limited  may  be  solved  thus: 

First,  we  find  the  gain  or  loss  on  one  unit  as  in  (730). 

Second,  we  balance  the  whole  gain  or  loss  on  an  ingredient 
where  the  quantity  is  limited,  by  using  any  ingredient  giving 
an  opposite  result  thus : 

Producing  Gain.  Gained  and  Lost.         Producing  Loss. 

(1.)  270  bu.  at  30  ct.  per  bu.  gain.  =  $81.00  =  405  bu.  at  20  ct.  per  ba.  loss. 
(2.)       2bu.at  loct.  perba.gam.=         .30=      3  bu.at  10  ct.per  bu.los8. 
272  bu.  +  408  bu.  =  680  bu.  in  mixture. 

Observe,  the  gain  on  the  370  bu.  may  be  balanced  with  the  other 
ingredient  that  produces  a  loss,  or  with  both  ingredients  that  produce  a 
loss,  and  these  may  be  put  in  the  mixture  in  different  proportions ; 
hence  a  series  of  different  mixtures  may  thus  be  formed. 

(313) 


186  BUSINESS    ARITHMETIC. 

6.  A  merchant  having  good  flour  worth  $7,  $9,  and  $12  a 
barrel,  and  240  barrels  of  a  poorer  quality  worth  $5  a  barrel, 
wishes  to  sell  enough  of  each  kind  to  realize  an  average  price 
of  110  a  barrel  on  the  entire  quantity  sold.  How  many  bar- 
rels of  each  kind  can  he  sell  ? 

7.  I  wish  to  mix  vinegar  worth  18,  21,  and  27  cents  a 
gallon  with  8  gallons  of  water,  making  a  mixture  worth 
25  cents  a  gallon  ;  how  much  of  each  kind  of  vinegar  can  I  use  ? 

8.  A  man  bought  a  lot  of  sheep  at  an  average  price  of  $2 
apiece.  He  paid  for  50  of  them  $2.50  per  head,  and  for  the 
rest  $1.50,  $1.75,  and  $3.25  per  head;  how  many  sheep  could 
there  be  in  the  lot  at  each  price  ? 

9.  A  milkman  mixes  milk  worth  8  cents  a  quart  with 
water,  making  24  quarts  worth  6  cents  a  quart ;  how  much 
water  did  he  use  ^ 

Examples  like  this,  where  the  quantity  of  the  mixture  is 
limited,  may  be  solved  thus : 

Solution.— 1.  We  find,  according  to  (720),  the  smallest  proportional 
parts  that  can  be  used,  namely,  3  quarts  of  milk  and  1  quart  of  water, 
making  a  mixture  of  4  quarts. 

2.  Now,  since  in  4  qt.  of  the  mixture  there  are  3  qt.  of  milk  and  1  qt. 
of  water,  in  24  qt.  there  must  be  as  many  times  3  qt.  of  milk  and  1  qt.  of 
water  as  4  qt.  are  contained  times  in  24  qt.  Consequently  we  have  as  the 
first  step  24  qt.  -f-  4  qt.  =  6,  second  step  3  qt.  x  6  =  18  qt.  and  1  qt.  x  6 
=  G  qt.  Hence  in  24  qt.  of  the  mixture  there  are  18  qt.  of  milk  and 
6  qt.  of  water. 

10.  A  grocer  has  four  kinds  of  coffee  worth  20,  25,  35,  and 
40  cents  a  pound,  from  which  he  fills  an  order  for  135  pounds 
worth  32  cents  a  pound ;  how  may  he  form  the  mixture  ? 

11.  A  jeweler  melts  together  gold  14,  18,  and  24  carats  fine, 
so  as  to  make  240  oz.  22  carats  fine ;  how  much  Of  each  kind 
did  it  require  ? 

12.  I  wish  to  fill  an  order  for  224  lb.  of  sugar  at  12  cents, 
by  forming  a  mixture  from  8,  10,  and  IG  cent  sugar  ;  how 
much  of  each  must  I  take  ? 

(314) 


3- 


-"-^^^ 


DEFINITIONS. 


723.  A  Power  of  a  number  is  either  the  number  itself 
or  the  product  obtained  by  taking  the  number  two  or  more 
times  as  a  factor. 

Thus  25  is  the  product  of  5  x  5  or  of  5  taken  twice  as  a  factor ;  hence 
25  is  a  power  of  5. 

724.  An  Exponent  is  a  number  written  at  the  right 
and  a  Uttle  above  a  number  to  indicate  : 

(1.)  The  number  of  times  the  given  number  is  taken  as  a  factor.  Thus 
in  7^  the  3  indicates  that  the  7  is  taken  3  times  as  a  factor ;  hence 
73  =  7x7x7  =  343. 

(2.)  The  degree  of  the  powor  or  the  order  of  the  power  with  reference 
to  the  other  powers  of  the  given  number.  Thus,  in  5"*  the  4  indicates 
that  the  given  power  is  the  fourth  power  of  5,  and  hence  there  are  three 
powers  of  5  below  5^ ;  namely,  5,  5"^  and  5^. 

735.  The  Square  of  a  number  is  its  second  power,  so 
called  because  in  finding  the  superficial  contents  of  a  given 
square  we  take  the  second  power  of  the  number  of  linear  units 
in  one  of  its  sides  (404). 

736.  The  Cube  of  a  number  is  its  third  poiuer,  so  called 
because  in  findmg  the  cubic  contents  of  a  given  cube  we  take 
the  tliird  poiver  of  the  number  of  linear  units  in  one  of  its 
edges  (413). 

737.  Involution  is  the  process  of  finding  any  required 
power  of  a  given  number. 

(315) 


188  BUSIITJSSS    ARITHMETIC, 

PROBLEMS  IN  INVOLUTION. 

728.  Pkob.  I.  —  To  find  any  power  of  any  given 
number. 

1.  Find  the  fourth  power  of  17. 

Solution. — Since  according  to  (721)  the  fourth  power  of  17  is  the 
product  of  17  taken  as  a  factor  4  times,  we  have  17  x  17  x  17  x  17=83521, 
the  required  power. 

2.  Find  the  second  power  of  48.     Of  65.    Of  432. 

3.  Find  the  square  of  294.    Of  386.     Of  497.     Of  253. 

4.  Find  the  cube  of  63.     Of  25.     Of  76.     Of  392. 

5.  Find  the  third  power  of  4.    Of  |.     Of  ^.     Of  .8. 

Observe,  any  power  of  a  fraction  is  found  by  involving  each  of  its 
terms  separately  to  the  required  power  (267). 

Find  the  required  power  of  the  following : 

6.  2372.      8.  (iJ)3.       10.  (.25)4.      12.  (.7f)2.      14.  (.005J)8. 

7.  45^.        9.  (^)^    11.  (.3|)3.      13.  (.l^^y.    15.  .03022. 

739.  Prob.  II.— To  find  the  exponent  of  the  pro- 
duct of  two  or  more  powers  of  a  given  number. 

1.  Find  the  exponent  of  product  of  7^  and  7^. 

Solution.— Since  7^  =  7  x  7  x  7  and  7^  =  7  x  7,  the  product  of  7'  and 
72  must  be  (7  X  7  X  7)  X  (7  X  7),  or  7  taken  as  a  factor  as  many  times  as  the 
sum  of  the  exponents  3  and  2.  Hence  to  find  the  exponent  of  the  pro- 
duct of  two  or  more  powers  of  a  given  number,  we  take  the  sum  of  the 
given  exponents. 

Find  the  exponent  of  the  product 

2.  Of  35*  X  353.  4.  Of  182  X  181  6.  Of  23^  x  235. 

3.  Of  (1)5  X  (1)2.     5.  Of  (i)7  X  (i)«.      7.  Of  (^Y  X  i^y. 

8.  Of  (74)2.     Observe,  (7^f  =  7*  x  7*  =  7*^2 :,:,  78. 
Hence  the  required  exponent  is  the  product  of  the  given  exponents. 

9.  Of  (123)4.      10.  Of  (96)5.      11.  Of  (168)8.      12.  Of  [(iYY. 

(316) 


£y^^ 


[e  V  O  L.  U  T  I  O  N]]  ^(^ 
.» ,— ^^„— ^ .-. ®> 


DEFINITIONS. 

730.  A  Itoot  of  a  number  is  either  the  number  itself-  op 
one  of  the  equal  factors  into  which  it  can  be  resolved. 
Thus,  since  7  x  7  =  49,  the  factor  7  is  a  root  of  49. 

-    731.   The  Second  or  Square  Hoot  is  one  of  the  two 

equal  factors  of  a  number.     Thus,  5  is  the  square  root  of  35. 

732.  The  Tliird  or  Cube  Hoot  is  one  of  the  three 
equal  factors  of  a  number.     Thus,  2  is  the  cube  root  of  8. 

733.  The  Madical  or  Hoot  Sign  is  ^,  or  ^fractional 
exponent. 

Wlien  the  sign,  |/,  is  used,  the  degree  or  name  of  the  root  is  indicated 
by  a  small  figure  written  over  the  sign  ;  when  the  fractional  exponent  is 
used,  the  denominator  indicates  the  name  of  the  root ;  thus, 

^9  or  9"  indicates  that  the  second  or  square  root  is  to  be  found. 

Y^27  or  27^  indicates  that  the  third  or  cube  root  is  to  be  found. 
Any  required  root  is  expressed  in  the  same  manner.    The  index  is 
usually  omitted  when  the  square  root  is  required. 

734.  A  Perfect  Power  is  a  number  whose  exact  root 
can  be  found. 

735.  An  Imperfect  Power  is  a  number  whose  exact 
root  cannot  be  found. 

The  indicated  root  of  an  imperfect  power  is  called  a  surd  ;  thus  ^/^. 

736.  Evolution  is  the  process  of  finding  the  roots  of 
numbers. 

(317) 


190 


BUSIjyUSS    ARITHMETIC, 


SQUARE  ROOT. 

PBEPABATOltT     rJtOPOSITIONS, 

737.  Prop.  I. — Any  perfect  second  poioer  may  he 
represented  to  the  eye  by  a  square,  and  the  number  of  units  in 
the  side  of  such  square  luill  represent  the  second  or  square 
ROOT  of  the  given  power. 

For  example,  if  25  is  the  given  power,  we  can  suppose  the 
number  represents  25  small  squares  and  arrange  them  thus : 

1.  Since  25  =  5  x  5,  we  can  arrange  the  25  squares 
5  in  a  row,  making  5  rows,  and  hence  forming  a  square 
as  shown  in  the  illustration. 

2.  Since  the  side  of  the  square  is  5  units,  it  represents 
the  square  root  of  25,  the  given  power;  hence  the  truth  of  the 
proposition. 

738.  Prop.  II. — Any  number  being  given,  by  supposing  it 
to  represent  small  squares,  we  can  find  by  arranging  these 
squares  in  a  large  square  the  largest  perfect  second  power  the 
given  number  contains,  and  hence  its  square  root. 

For  example,  if  we  take  83  as  the  given  number  and  sup- 
pose it  to  represent  83  small  squares,  we  can  proceed,  thus : 


'"1  ;.[■■, 

■ 

.1 

_■„ 

■1 

_!!: 

L. 

(1) 


(2) 


1.  We  can  take  any  number  of  the  83  squares,  as  36, 
that  we  know  will  form  a  perfect  square  (Prop.  I), 
and  arrange  them  in  a  square,  as  shown  in  (I),  leav- 
ing 47  of  the  83  squares  yet  to  be  disposed  of. 

2.  We  can  now  place  a  row  of  squares  on  two 
adjacent  sides  of  the  square  in  (1)  and  a  square  in  the 
corner,  and  still  have  a  perfect  square  as  shown  in  (2\ 

3.  Observe,  that  in  putting  one  row  of  small  squares  r:i 
each  of  two  adjacent  sides  of  the  square  first  formed, 
we  must  use  twice  as  many  squares  as  there  are  units 
in  the  side  of  the  square. 

4.  Now  since  it  takes  twice  6  or  12  squares  to  put 
one  row  on  each  of  two  adjacent  sides,  we  can  put  on  as  many  rows  as 

(318) 


I  I  I  I  I  W 


EVOLUTION-. 


191 


6xr 

(3) 
=18. 

3^=9. 

1 

1 

■ 

1 

1 

1 

---1 

a 

f  -       ^-:- 

^1: 

tj-- 

c 

a 

12  is  contained  times  in  47,  the  number  of  squares  remaining.    Hence 
we  can  put  on  3  rows  as  shown  in  (3)  and  have  11  squares  still  remaining. 

5.  Again,  having  put  3  rows  of  squares  on  each  of 
two  adjacent  sides,  it  takes  3  x  3  or  9  squares  to  fill 
the  comer  thus  formed,  as  shown  in  (3),  leaving  only 
2  of  the  1 1  squares. 

Hence,  the  square  in  (3)  represents  the  greatest 
perfect  power  in  83,  namely  81 ;  and  9,  the  number  of 
units  in  its  side,  represents  the  square  root  of  81. 

6.  Now  cbserw  that  the  length  of  the  side  of  the 
square  in  (3)  is  6  +  3  units,  and  that  the  number  of 

(Small  squares  may  be  represented  in  terms  of  6  +  3  ;  thus, 

(1.)  (6  +  3)2  =  Q^-]-^^-\-Uvice  6x3  =  36  +  9  +  36  =  81. 

Again,  suppose  5  units  had  been  taken  as  the  side  of  the  first  square, 
the  number  of  small  squares  would  be  represented  thus  : 

(2.)  (5  +  4)2  ^  52  +  42  +  ^mce  5  x4  z=  25  +  16  +  40  =  81. 

In  the  same  manner  it  may  be  shown  that  the  square  of  the  sum  of 
any  two  numbers  expressed  in  terms  of  the  numbers,  is  the  square  of 
each  of  the  numbers  plus  twice  i\\e\t  product. 

Hence  the  square  of  any  number  may  be  expressed  in  terms  of  its 
tens  and  units  ;  thus  57  =  50  +  7  ;  hence 

(3.)  572  =  (50  +  7)2  =  bOf^+'l^+tivice  50  x  7  =  3249. 

Tliis  may  also  be  shown  by  actual  multiplication.  Thus,  in  multiply- 
ing 57  by  57  we  have,  first,  57  x7  =  7x7  +  50x7  =  7^  +  50  x7;  we  have, 
second,  57  x  50  =  50  x  7  +  50  x  50  =  50  x  7  +  50^ ;  hence,  57^  =  50^  +  7'^  + 
twice  50  X  7. 

Find,  by  constructing  a  diagram  as  above,  the  square  root  of 
each  of  the  following  : 

Observe,  that  when  the  number  is  large  enough  to  give  tens  in  the 
root,  we  can  take  as  the  side  of  the  first  square  we  construct  the  greatest 
number  of  tens  whose  square  can  be  taken  out  of  the  given  number. 

1.  Of  144.  4.  Of  529.  7.  Of  1125.        10.  Of  1054. 

2.  Of  196.  5.  Of  729.  8.  Of  584.  11.  Of  2760. 

3.  Of  289.  6.  Of  1089.         9.  Of  793.  12.  Of  3832. 

(319) 


192  BUSINESS    ARITH3IETIC. 

739,  Prop.  III. — Tlie  square  of  any  mimher  must  con- 
tain twice  as  many  figures  as  the  number,  or  twice  as  many 
less  one. 

This  proposition  may  be  shown  thus : 

1.  Observe,  the  square  of  either  of  the  digits  1,  2,  3,  is  expressed  by 
one  figure,  and  the  square  of  either  of  the  digits  4,  5,  6,  7,  8,  9,  is 
expressed  by  two  figures;  thus,  2x3=4,  3x3  =  9,  and  4x4  =  16, 
5  X  5  =  25,  and  so  on. 

2.  Since  10  x  10  =  100,  it  is  evident  the  square  of  any  number  of  tens 
must  have  two  ciphers  at  the  right ;  thus,  20'  =  20  x  20  =  400. 

Now  since  the  square  of  either  of  the  digits  1,  2,  3,  is  expressed  by 
one  figure,  if  we  have  1,  2,  or  3  tens,  the  square  of  the  number  must  be 
expressed  by  3  figures ;  that  is,  one  figure  less  than  twice  as  many  as  are 
required  to  express  the  number. 

Again,  since  the  square  of  either  of  the  digits  4,  5,  6,  7,  8,  9,  is 
expressed  by  two  figures,  if  we  have  4,  5,  6,  7,  8,  or  9  tens,  the  square 
of  the  number  must  contain  four  figures  ;  that  is,  twice  as  many  figures 
as  are  required  to  express  the  number.  Hence  it  is  evident  that,  in  the 
square  of  a  number,  the  square  of  the  tens  must  occupy  the  third  or  the 
third  and  fourth  place. 

By  the  same  method  it  may  be  shown  that  the  square  of  hundreds 
must  occupy  the  fifth  or  the  fifth  and  sixth  places,  the  square  of 
thousands  the  seventh  or  the  seventh  and  eighth  places,  and  so  on  ;  hence 
the  truth  of  the  proposition. 

From  this  proposition  we  have  the  following  conclusions : 

740.  /.  If  any  numler  he  separated  into  periods  of  two 
figures  each,  beginning  with  the  units  place,  the  number  of 
periods  will  be  equal  to  the  number  of  places  in  the  square 
root  of  the  greatest  perfect  power  luhich  the  given  number 
contains. 

11.  In  the  square  of  any  number  the  square  of  the  units  are 
found  in  the  units  and  tens  place,  the  square  of  the  tens  in 
the  hundreds  and  thousands  place,  the  square  of  the  hun- 
dreds in  the  tens  and  hundreds  of  thousands  place, 
and  so  on. 

(320) 


EVOL  UTION, 


193 


IZZUSTJtATIOK     OF     PROCESS, 

741.     1.  Find  the  square  root  of  225. 

(c)  10x5=50.       (^5- =25. 


1st  Step. 
2d  Step. 


(a)  10^=100. 


225(10 
102  =  10x10=  100 

i(l)  T.  divisor  10  x  2=20)125  (  5 
root  15 

Explanation.  —  1.  We  observe,  as 
shown  in  (a),  that  1  ten  is  the  largest  num- 
ber of  tens  whose  square  is  contained  in 
225.  Hence  in  1st  step  we  subtract  10^= 
100  from  225,  leaving  125. 

2.  Having  formed  a  square  whose  side  is  10  units,  we  observe,  as 
shown  in  (&)  and  (<•),  that  it  will  take  twke  ten  to  put  one  row  on  two 
adjacent  sides.    Hence  the  Trial  Divisor  is  10  x  2=20. 

3.  We  observe  that  20  is  contained  6  times  in  125,  but  if  we  add  6  units 
to  the  side  of  the  square  {a)  we  will  not  have  enough  left  for  the  corner 
(d),  hence  we  add  5  units. 

4.  Having  added  5  units  to  the  side  of  the  square  (a),  we  observe,  as 
shown  in  (b)  and  (c),  that  it  requires  t^cice  10  or  20  multiplied  by  5  plus 
5  X  5,  as  shown  in  (d),  to  complete  the  square ;  hence  (2)  in  2d  step. 

Solution  with  every  Operation  Indicated^ 

743.    2.  Find  the  square  root  of  466489. 


FmsT  Step. 


600x600 


466489  (  600 
360000       80 


Second  Step, 


80x80: 


6400 )  ~  102400     683  required  root 


Thibd  Step, 


r  (1)  Trial  divisor  600  x  2=1200)106489 

•i       (1200x80=96000/ 
((2)] 

|(1)  3 
!(2)| 


^(1)  Trial  divisor  m)x2=\Zm)    4089 
•    \_  i  1360x3=4080/ 


3x3= 


r 


Explanation.— 1.  We  place  a  point  over  every  second  figure 
beginning  with  the  units,  and  thus  find,  according  to  (740),  that  the 
root  must  have  three  places.  Hence  the  first  figure  of  the  roofc 
expresses  hundreds. 

(321) 


194  BUSINESS    ARITHMETIC, 

2.  We  observe  that  the  square  of  600  is  the  greatest  second  power  of 
hundreds  contained  in  466489.  Hence  in  the  first  step  we  subtract 
600  X  600  =  360000  from  466489,  leaving  103489. 

3.  We  now  double  the  600,  the  root  found,  for  a  trial  divisor, 
according  to  (741 — 2).  Dividing  106489  by  1200  we  find,  according  to 
(741 — 2),  that  we  can  add  80  to  the  root.  For  this  addition  we  use,  as 
shown  in  (2),  second  step,  1200  x  80  =  90000  and  80  x  80  =6400  (741—3), 
making  in  all  102400.  Subtracting  102400  from  106489,  we  have  still 
remaining  4089. 

4.  We  again  double  680,  the  root  found,  for  a  trial  divisor,  according 
to  (741 — 2),  and  proceed  in  the  same  manner  as  before,  as  shown  in 
third  step. 


•743,  Contracted  Solution  of  the  foregoing  Example. 

466489  ( 683 
First  Step.  6x6=         36 


1064 


„  ^         I  (1)        6x2=  12 

Second  Step.  {  ^  ' 

\  (2)     128  X  8  =  1024 


THIRD  Step,    j^      68x2  =  136)    4089 
1(2)  1363x3  =  4089 

Explanation. — 1.  Obseru,  in  the  first  step  we  know  that  the  square 
600  must  occupy  the  fifth  and  sixth  place  (738).  Hence  the  ciphers  are 
omitted. 

3.  Observe,  that  in  (1),  second  step,  we  use  6  instead  of  600,  thus 
dividing  the  divisor  by  100;  hence  we  reject  the  tens  and  units  from  the 
right  of  the  dividend  (142). 

3.  Observe,  also,  in  (2),  second  step,  we  imite  in  one  three  operations. 
Instead  of  multiplying  12  by  80,  the  part  of  the  root  found  by  dividing 
1064  by  12,  we  multiply  the  12  first  by  10  by  annexing  the  8  to  it  (91 ), 
and  having  annexed  the  8  we  multiply  the  result  by  8,  which  gives  us 
the  product  of  12  by  80,  plus  the  square  of  8.  But  the  square  of  8, 
written,  as  it  is,  in  the  third  and  fourth  place,  is  the  square  of  80. 

Hence  by  annexing  the  8  and  writing  the  result  as  we  do,  we  have 
united  in  oTie  three  operations ;  thus,  128  x  8  =  12  x  80  +  80  x  80. 

(322) 


EVOLUTION,  195 

From  these  illnstrations  we  have  the  following 

744.  EuLE. — /.  Separate  the  nurriber  into  periods  of 
two  figures  each,  by  placing  a  point  over  every  second 
figure,  beginning  with  the  units  figure. 

II.  Find  the  greatest  square  in  the  left-hand  period 
and  place  its  root  on  the  right.  Subtract  this  square 
froiiv  the  period  and  annex  to  the  reinainder  the  next 
period  for  a  dividend. 

III.  Double  the  part  of  the  root  found  for  a  trial 
divisor,  and  find  how  many  times  this  divisor  is  con- 
tained in  the  dividend,  omitting  the  right-hand  figure. 
Annex  the  quotient  thus  found  both  to  the  root  and  to 
the  divisor.  Multiply  the  divisor  thus  completed  by  the 
figure  of  the  root  last  obtained,  and  subtract  the  product 
froin  the  dividend. 

IV.  If  there  are  more  periods,  continue  the  operation 
in  the  same  manner  as  before. 

In  appl3^ng  this  rule  be  particular  to  observe 

1.  When  there  is  a  remainder  after  the  last  period  has  been  used, 
annex  periods  of  ciphers  and  continue  the  root  to  as  many  decimal 
places  as  may  be  required. 

2.  We  separate  a  number  into  periods  of  two  figures  by  beginning  at 
the  units  place  and  proceeding  to  the  left  if  the  number  is  an  integer, 
and  to  the  right  if  a  decimal,  and  to  the  right  and  left  if  both. 

3.  Mixed  numbers  and  fractions  are  reduced  to  decimals  before 
extracting  the  root.  But  in  case  the  numerator  and  the  denom- 
inator are  perfect  powers,  or  the  denominator  alone,  the  root  may  bo 
more  readily  formed  by  extracting  the  root  of  each  term  separately, 

rru      .  /49       1/49       7        ,       /35       ^35       V35 

so  on. 
Extract  the  square  root 

1.  OfVW-  3.  Of  HI.         5.  Of  J||.  7.  Of^^. 

2.  Of^.  4.  OffH-         6.  Oft^.  8.  Of^ftV- 

(323) 


196'  BUSINESS    ARITHMETIC. 

EXAMPLES     FOR     PRACTICE. 

745.  Extract  the  square  root 

1.  Of  4096.  7.  Of^^.  13.  Of  137641. 

2.  Of  3481.  8.  Offifi^.  14.  Of  4160.25. 

3.  Of  2809.  9.  Of  ttf  f  15.  Of  768427.56. 

4.  Of  7569.  10.  Of  .0225.  16.  Of  28022.76. 

5.  Of  8649.  11.  Of  .2304.  17.  Of  57.1536. 

6.  Of  9216.  12.  Of  .5776.  18.  Of  474.8041. 

Find  the  square  root  to  three  decimal  places 

19.  Of  32.  22.  Of  .93.  25.  Of  14.7.  28.  Of  J. 

20.  Of  59.  23.  Of  .8.  26.  Of  86.2.  29.  Of  ^V 

21.  Of  7.  24.  Of  .375.        27.  Of  5.973.        30.  Of  A- 
Perform  the  operations  indicated  in  the  following : 

31.  \/6889  — 'v/I024.  34.  76796^-^^2136. 

32.  V2209  + a/225.  35.  \/558009-^(TjJ:r)"^. 

33.  VfHl  X  a/2209.  36.  (f f f|)^  X 131376^. 

37.  What  is  the  length  of  a  square  floor  containing  9025 
square  feet  of  lumber  ? 

38.  A  square  garden  contains  237169  square  foot;  how 
many  feet  in  one  of  its  sides  ? 

39.  How  mony  yards  in  one  of  the  equal  sides  of  a  square 
acre? 

40.  An  orchard  containing  9216  trees  is  planted  in  the 
form  of  a  square,  eacli  tree  an  equal  distance  from  another ; 
how  many  trees  in  each  row  ? 

41.  A  triangular  field  contains  1966.24  P.  What  is  the 
length  of  one  side  of  a  square  field  of  equal  area  ? 

42.  Find  the  square  root  of  2,  of  5,  and  of  11,  to  4  decimal 
places. 

43.  Find  the  square  root  of  4,  ^,  and  of  |},  to  3  decimal 
places. 

(324) 


VOL  UTION. 


m 


CUBE  ROOT. 


PJREPAJtATOIiT     PROPOSITIONS, 

46.  Prop.  I. — A  ny  perfect  third  power  may  he  repre- 
sented  to  the  eye  hy  a  cube,  mid  the  number  of  units  in  the  side 
of  such  cube  will  represent  the  third  or  cube  root  of  the 
given  power. 

Represent  to  the  eye  by  a  cube  343. 

(1)  1.  We  can  suppose  the  number  343  to  repre- 

sent small  cubes,  and  we  can  take  2  or  more  of 
these  cubes  and  arrange  tliem  in  a  row,  as  shown 
in  (1). 

2.  Having  formed  a  row  of  5  cubes,  as  shown  in  (1),  we  can  arrange 

6  of  these  rows  side  by  side,  as  shown  in  (3),  forming  a  square  slab 

containing  5x5  small  cubes,  or  as  many  small 

("^)  cubes  as  the  square  of  the  number  of  unUs  in  the 

side  of  the  slab. 

3.  Placing  5  such  slabs  together,  as  shown  in 
(3),    we    form    a    cube.     Now,    since  each    slab 

contains  5x5  small  cubes,  and  since  5  slabs  are  placed  together,  the 
cube  in  (3)  contains  5  x  5  x  5,  or  125  small  cubes,  and  hence  represents 
the  tldvd  power  125,  and  each  edge  of  the  cube 
^^^  represents  to  the  eye  5,  the  cube  root  of  125. 

We  have  now  remaining  yet  to  be  disposed  of 
843—125,  or  218  small  cubes. 

4.  Now,  ohserce,  that  to  enlarge  the  cube  in  (3) 
so  that  it  may  contain  the  343  small  cubes,  we 
must  build  the   same  number  of  tiers  of  small 

cubes  upon  each  of  three  adjacent  sides,  as  shown  in  (4). 

Obnerve,  also,  that  a  slab  of  small  cubes  to  cover  one  side  of  the  cube 
in  (3)  must  contain  5x5  or  25  small  cubes,  as 
shown  in  (4),  or  as  many  small  cubes  as  the  square 
of  the  number  of  units  in  one  edge  of  the  cube 
in  (3). 

Hence,  to  find  the  number  of  cubes  necessary 
to  put  one  slab  on  each  of  three  sides  of  the  cube 
in  (3),  we  nmltiply  the  square  of  its  edge  by  3 
giving  5'  X  3  =  5  X  5  X  3  =  75  small  cubes. 
(325) 


198 


B  USINESS    ARITHMETIC. 


Examine  careful 


5.  Having  found  that  75  small  cubes  will  put  one  tier  on  each  of 
three  adjacent  sides  of  the  cube  in  (3),  we  divide  218,  the  number  of 
small  cubes  yet  remaining,  by  75,  and  find  how 
many  such  tiers  we  can  form.  Thus,  218-^75  = 
2  and  68  remaining.  Hence  we  can  put  2  tiers 
on  each  of  three  adjacent  sides,  as  shown  in  (5), 
and  have  G8  small  cubes  remaining. 

G.  Now,  observe,  that  to  complete  this  cube 
we  must  fill  each  of  the  thj'ee  corners  formed  by 
building  on  three  adjacent  sides. 
(6)  and  obserce  that  to  fill  one  of  these  three  corners 
we  require  as  many  small  cubes  as  is  expressed 
by  the  square  of  the  number  of  tiers  added, 
multiplied  by  the  number  oi  units  in  the  side  of 
the  cube  to  which  the  addition  is  made.  Hence 
we  require  2^  x  5  or  20  small  cubes.  And  to  fill 
the  three  corners  we  require  3  times  2"^  x  5  or  60, 
leaving  68—60  or  8  of  the  small  cubes. 

7.  Examine  again  (5)  and  (6)  and  observe  that 
when  the  three  comers  are  filled  we  require  to  complete  the  cube  as 
^~^  shown  in  (7),  another  cube  whose  side  contains  as 

many  units  as  there  are  units  added  to  the  side  of 
the  cube  on  which  we  have  built.  Consequently 
we  require  2^  or  2  x  2  x  2  =  8  small  cubes. 

Hence   we    have  formed  a   cube    containing 
343  small  cubes,  and  any  one  of  its  edges  repre- 
sents to  the  eye  5  +  2  or  7  units,  the  cube  root  of  343. 
From  these  illustrations  it  will  be  seen  that  the 
Bteps  in  finding  the  cube  root  of  343  may  be  stated  thus  : 
We  assume  that  343  represents  small  cubes,  and  take 
5  as  the  length  of  the  side  of  a  large  cube  formed 
from  these.    Hence  we  subtract  the  cube  of  5  = 
f  1.  We  observe  it  takes  5^  x  3  =  75  to  put  one  tier  on 
three  adjacent  sides.    Hence  we  can  put  on 
2.  We  have  now  found  that  we  can  add  2  units  to  the 
side  of  the  cube.  Hence  to  add  this  we  require 

(1)  For  the  3  sides  of  the  cube     5^  X  2  X  3  =  150  ^ 

(2)  For  the  3  corners  thus  formed  2^x5x3=    60  >=  218 
.  (3)  For  the  cube  in  the  comer  last  formed  2^=      8  J 

Hence  the  cube  root  of  343  is  5  +  2  =  7. 

(326) 


FmsT  Step. 


Second  Step. 


343 

125 

75)218(2 


E  VOL  UTION. 


199 


747.  Ohserm,  that  the  number  of  small  cubes  in  the  cube  (7)  in  the 
foregoinr:  illustrations,  are  expressed  in  terms  of  5  +  3  ;  namely,  the  num- 
ber of  units  in  the  side  of  the  first  cube  formed,  plus  the  number  of  tiers 
added  in  enlarging  this  cube  ;  thus : 


(5  +  2)^ 


53+52x2x3      +     22x5x3  +  21 


In  this  manner  it  may  be  shown  that  the  cube  of  the  sum  of  any  two 
numbers  is  equal  to  the  cube  of  each  number,  plus  3  times  the  square  of 
the^r*^  multiplied  by  the  second  number,  plus  3  times  the  square  of  the 
second  multiplied  by  the^r«^  number. 

Hence  the  cube  of  any  number  may  be  expressed  in  terms  of  its  tens 
and  units  ;  thus,  74  =  70  +  4 ;  hence, 

(70  +  4)3  =703  +  3  times  70^  x  4  +  3  times  4^  x  70  +  43  =  405224. 

Solve  each-  of  the  following  examples,  by  applying  the  fore- 
going illustrations : 

1.  Find  the  side  of  a  cube  which  contains  729  small  cubes, 
taking  6  units  as  the  side  of  the  first  cube  formed. 

2.  Take  20  units  as  the  side  of  the  first  cube  formed,  and 
find  the  side  of  the  cube  that  contains  15625  cubic  units. 

3.  How  many  must  be  added  to  9  that  the  sum  may  be  the 
cube  root  of  4096?    Of  2197?     Of  2744? 

4.  Find  the  cube  root  of  1368.    Of  3405.   Of  2231.    Of  5832. 

5.  Express  the  cube  of  83  in  terms  of  80  +  3. 

6.  Express  the  cube  of  54,  of  72,  of  95,  of  123,  of  274,  in 
terms  of  the  tens  and  units  of  each  number. 

(327) 


200  BUSINESS    ARITHMETIC, 

748.  Pkop.  II. — The  cube  of  any  member  must  contain 
three  times  as  many  places  as  the  number,  or  three  times  as 
many  less  one  or  two  places. 

This  proposition  may  be  shown  thus : 

1.  Observe,  1^  =  1,  2^  =  9,  3^  =  27,  ¥  =  64,  5^  =  125,  and  9^  =  729  ; 
hence  the  cube  of  1  and  2  is  expressed  each  by  one  figure,  the  cube  of  3 
and  4  each  by  two  figures,  and  any  number  from  5  to  9  inclusive  each 
by  three  figures. 

2.  Observe,  also,  that  for  every  cipher  at  the  right  of  a  number  there 
must  (91)  be  three  ciphers  at  the  right  of  its  cube;  thus,  10^  =  1,000, 
100"^  =  1,000,000.  Hence  the  cube  of  tens  can  occupy  no  place  lower 
than  thousands,  the  cube  of  hundreds  no  place  lower  than  mUlicns,  and 
so  on  with  higher  orders. 

8.  From  the  foregoing  we  have  the  following : 

(1.)  Since  the  cube  of  1  or  2  contains  one  figure,  the  cube  of  1  or  2  tens 
must  contain /6>wr  places;  of  1  or  2  hundreds^  seven  places,  and  so  on 
with  higher  orders. 

(2.)  Since  the  cube  of  3  or  4  contains  two  figures,  the  cube  of  3  or  4  tens 
must  contain  jive  places  ;  of  3  or  4  hundreds,  eight  places,  and  so  on  with 
higlier  orders. 

(3.)  Since  the  cube  of  any  number  from  5  to  9  inclusive  contains 
three  places,  the  cube  of  any  number  of  tens  from  5  to  9  tens  inclusive 
must  contain  six  places  ;  of  hundreds,  from  5  to  9  hundred  inclasive,  nine 
places,  and  so  on  with  higher  orders ;  hence  the  truth  of  the  propos'tion. 

Hence  also  the  following  : 

749.  /.  If  any  number  be  separated  into  periods  of  three 
figures  each,  beginning  with  the  units  place,  the  number  of 
periods  will  be  equal  to  the  number  of  places  in  the  cube  root 
of  the  greatest  perfect  third  power  which  the  given  mtmhcr 
contains. 

II.  The  cube  of  units  contains  no  order  higher  than 
hundreds. 

III.  TJie  cube  of  tens  contains  no  order  lower  than  thousands 
nor  higher  than  hundred  thousands,  the  cube  of  hundreds  no 
order  lower  than  millions  nor  higher  than  hundred  millions^ 
and  so  on  with  higher  orders. 

(328) 


EVOLUTION,  201 

IZrUaTJtATION     OF     mOC  ESS. 

150.  Solution  with  every  Operation  Indicated, 

Find  the  cube  root  of  92345408. 

92345408  (  400 
First  Step.  400^=400  x  400  x  400  =  64000000 


Second  Step. 


Third  Step. 


(1)  Trial  divisor  400-  x  3=480000  )  28345408  (    50 

4002  X  50x3  =  24000000 \ 

(2)  ^  50-'  X  400  X  3  =   3000000  V   =  27125000 
503=      135000 ) 


(1)  Trial  divisor  450^  x  3=607500  )    1220408  (      2 

) 


450^x3x2  =1215000)  Moot  462 

(2)  -J    2*  X  450x3=      5400^  1220408 


2»=  8  J 

Explanation. — 1.  We  place  a  period  over  every  third  figure  begin- 
ning with  the  units,  and  thus  find,  according  to  (749),  that  the  root 
must  have  three  places.  Hence  the  first  figure  of  the  root  expresses 
hundreds. 

2.  We  observe  that  400  is  the  greatest  number  whose  cube  is  contained 
in  the  given  number.  Subtracting  400^  =  64000000  from  92345408,  we 
have  28345408  remaining. 

3.  We  find  a  trial  divisor,  according  to  (746 — 4),  by  taking  3  times  the 
square  of  400,  as  shown  in  (1),  second  step.  Dividing  by  this  divisor, 
according  to  (746 — 5),  we  find  we  can  add  50  to  the  root  already  found. 

Observe,  the  root  now  found  is  400  +  50,  and  that  according  to  (747), 

(400  +  50)3  =  4003  +  4002  x  50  x  3  +  50'  x  400  x  3  +  503. 

We  have  already  subtracted  4003  =  640000  from  the  given  number. 
Hence  we  have  only  now  to  subtract, 

400'  X  50  X  3  +  502  x  400  X  3  +  503  ^  27125000, 

as  shown  in  (2),  second  step,  leaving  1220408. 

5.  We  find  another  trial  divisor  and  proceed  in  the  same  manner  to  find 
the  next  figure  of  the  root,  as  shown  in  the  third  step. 

(329) 


202 


BUSINESS     ARITHMETIC. 


751.  Contracted  Solution  of  the  foregoing  Example. 


First  Step. 


Second  Step. 


4^  =  4x4x4  = 

(1)  Trial  divisor  40'  x  3  = 

r  402  X  5  X  3  =  24000  ) 

(3)  <5-'x40x3=  3000  [  = 

(      53  =   125  3 


92345403  (  452 
G4 


4800  )  28345 
27125 


Third  Step. 


( (1)  Trial  divisor  450^ xS  =  6075( 

]         C  450-'  X  2  x  3  :=  1215000  ) 
((2)    ^2^x450x3=       5400>  = 


;  (1)  Trial  divisor  450^  x  3 

(       23  =     s) 


607500  )  1220408 
1220408 


Explanation.— 1.  Observe,  in  the  first  step,  we  know  that  the  cube 
of  400  must  occupy  the  seventh  and  eighth  places  (749— III).  Hence 
the  ciphers  are  omitted. 

2.  Observe,  also,  that  no  part  of  the  cube  of  hundreds  and  tens  is 
found  below  thousands  (749 — HI).  We  therefore,  in  finding  the  number 
of  tens  in  the  root,  disregard,  as  shown  in  second  step,  the  right-hand 
period  in  the  given  number,  and  consider  the  hundreds  and  tens  in  the 
root  as  tens  and  units  respectively. 

Hence,  in  general,  whatever  number  of  places  there  are  in  the  root, 
we  disregard,  in  finding  any  figure,  as  many  periods  at  the  right  of  the 
given  number  as  there  are  places  in  the  root  at  the  right  of  the  figure 
we  are  finding,  and  consider  the  part  of  the  root  found  as  tens,  and  the 
figure  we  are  finding  as  units,  and  proceed  accordingly. 

From  these  illustrations  we  have  the  following : 

*li52.    Rule. — I.  Separate  the  nmnherinto  periods  of  • 
three  figures  each,  hy  placing  a  point  over  every  third 
figure,  beginning  with  the  units  figure. 

II.  Find  the  greatest  cube  in  the  left-hand  period, 
and  place  its  root  on  the  right.  Subtract  this  cube  from 
the  period  and  annex  to  the  remainder  the  next  period 
for  a  dividend. 

III.  Divide  this  dividend  by  the  tHal  divisor,  which 
is  3  times  the  square  of  the  root  already  found,  con- 

(330) 


EVOLUTION,  203 

sidered  as  tens}   the  quotient  is  the  next  figure  of 
the  root. 

IV.  Subtract  from  the  dividend  S  times  the  square 
of  the  root  before  found,  considered  as  tens,  multiplied 
by  the  figure  last  found,  plus  3  times  the  square  of  the 
figure  last  found,  multiplied  by  the  root  before  found, 
plus  the  cube  of  the  figure  last  found,  and  to  the  re- 
mainder annex  the  next  period,  if  any,  for  a  new 
dividend. 

V.  If  there  are  more  figures  iiv  the  root,  find  in  the 
Same  manner  trial  divisors  and  proceed  as  before. 

In  applying  this  rule  be  particular  to  observe : 

1.  In  dividing  by  the  Trial  Bicisor  the  quotient  may  be  larger  than 
the  required  figure  in  the  root,  on  account  of  the  addition  to  be  made, 
as  shown  in  (T4C$ — 0)  mcond  step.  In  such  case  try  a  figure  1  less 
than  the  quotient  found. 

2.  When  there  is  a  remainder  after  the  last  period  has  been  used, 
annex  periods  of  ciphers  and  continue  the  root  to  as  many  decimal 
places  as  may  be  required. 

3.  We  separate  a  number  into  periods  of  three  figures  by  beginning 
at  the  units  place  and  proceeding  to  the  left  if  the  number  is  nn  integer, 
and  to  the  right  if  a  decimal,  and  to  the  right  and  left  if  both. 

4.  Mixed  numbers  and  fractions  are  reduced  to  decimals  before 
extracting  the  root.  But  in  case  the  mimerator  and  the  denominator 
are  perfect  third  powers,  or  the  denominator  alone,  the  root  may  be  more 
readily  found  by  extracting  the  root  of  each  term  separately. 

EXAMPIiES     FOR     PRACTICE. 

753.  Find  the  cube  root  of 

9.  fiff.     13.  24137569. 
10-  t¥^-     14.  47245881. 

11.  250047.         15.  tAtVW 

12.  438976.         16.  113.379904. 
17.  Find,  to  two  decimal  places,  the  cube  root  of  11.   Of  36. 

Of  84.     Of  235.     Of^.     Oi^-i^.     Of  75.4.     Of  6.7. 

(331) 


1.  216. 

5.  4096. 

2.  729. 

6.  10648, 

3.  1331. 

7.  6859. 

4.  2197. 

8.  A4V 

304  BUSINESS     ARITHMETIC. 

18.  Find  to  three  decimal  places  the  cube  root  of  3.  Of  7. 
Of  .5.     Of  .04.     Of  .009.     Of  2.06. 

19.  Find  the  sixth  root  of  4096. 

Observe,  the  sixth  root  may  be  found  by  extracting  first  the  square 
root,  then  the  cube  root  of  the  result. 

For  example,  y^4096  =  64  ;  hence,  4096=64  x  64.  Now,  if  we  extract 
the  cube  root  of  64  we  will  have  one  of  the  three  equal  factors  of  64, 
and  hence  one  of  the  six  equal  factors  or  sixth  root  of  4096. 

Thus,  ^^64  =4  ;  hence,  64  =  4x4x4.  But  we  found  by  extracting 
it3  square  root  that  4096  =  64  x  64,  and  now  by  extracting  the  cube  root 
that  64=  4x4x4;  consequently  we  know  that  4096 =(4  x  4  x  4)  x  (4  x  4  x  4). 
Hence  4  is  the  required  sixth  root  of  4096. 

In  this  manner,  it  is  evident,  we  can  find  any  root  whose  index  con- 
tains no  other  factor  than  2  or  3. 

20.  Find  the  sixth  root  of  2565726409. 

21.  Find  the  eighth  root  of  43046721. 

22.  What  is  the  fourth  root  of  34012224? 

23.  What  is  the  ninth  root  of  134217728? 

24.  A  pond  contains  84604519  cubic  feet  of  water;  what 
must  be  the  length  of  the  side  of  a  cubical  reservoir  which  will 
exactly  contain  the  same  quantity  ? 

25.  What  is  the  length  of  the  inner  edge  of  a  cubical  cistern 
that  contains  2079  gal.  of  water? 

^6.  How  many  square  feet  in  the  surface  of  a  cube  whose 
Tolume  is  16777216  cubic  inches? 

27.  A  pile  of  cord  wood  is  256  ft.  long,  8  ft.  high,  and  16  ft. 
wide;  what  would  be  the  length  of  the  side  of  a  cubical 
pile  containing  the  same  quantity  of  wood  ? 

28.  What  is  the  length  of  the  inner  edge  of  a  cubical  bin 
that  contains  3550  bushels  ? 

29.  What  are  the  dimensions  of  a  cube  whose  volume  is 
equal  to  82881856  cubic  feet? 

30.  What  is  the  length  in  feet  of  the  side  of  a  cubical 
reservoir  which  contains  1221187^  pounds  avoirdupois,  pure 
water? 

(332) 


1    ,.     -^rm>^    ,.,    1 

JT I  PROGRESSIONS  f0- 

DEFINITIONS. 

754.  A  ^Progression  is  a  series  of  numbers  so  related, 
that  each  number  in  the  series  may  be  found  in  the  same 
manner,  from  the  number  immediately  preceding  it. 

^55,  An  Arithmetical  Progression  is  a  series  of 
numbers,  which  increases  or  decreases  in  such  a  manner  that 
the  difference  between  any  Uvo  consecutive  numbers  is 
constant.     Thus,  3,  7,  11, 15,  19,  23. 

115^,  A  Geometrical  Progression  is  a  series  of 
numbers,  which  increase  or  decrease  in  such  a  manner  that 
the  ratio  between  any  tiuo  consecutive  numbers  is  constant. 

Thus,  5,  10,  20,  40,  80,  is  a  geometrical  progression. 

^511.  The  Terms  of  a  progression  are  the  numbers  of 
which  it  consists.  The  First  and  Last  Terms  are  called  the 
Extremes  and  the  intervening  terms  the  Means, 

^5S.  The  Common  or  Constant  JDifference  of  an 

arithmetical  progression  is  the  difference  between  any  two 
consecutive  terms. 

159.  The  Common  or  Constant  Matio  or  Multi- 
plier  of  a  geometrical  progression  is  the  quotient  obtained 
by  dividing  any  term  by  the  preceding  one. 

760.  An  Ascending  or  Increasing  Progression 

is  one  in  which  each  term  is  greater  than  the  preceding  one. 

761.  A  Descending  or  Decreasing  Progression 

is  one  in  which  each  term  is  less  than  the  preceding  one. 

(333) 


206 


BUSINESS    ARITHMETIC. 


ARITHMETICAL  PROGEESSION. 

762.  There  are  five  quantities  considered  in  Arithmetical 
Progression,  which,  for  convenience  in  expressing  rules,  we 
denote  by  letters,  thus : 

1.  A  represents  the  First  Term  of  a  progression. 

2.  JL  represents  tlie  Last  Term. 

3.  D  represents  tlie  Constant  or  Common  Difference, 

4.  N  represents  the  Numler  of  Terms. 

5.  S  represents  the  Sum  of  oM  the  Terms. 

763.  Any  three  of  these  quantities  being  given,  the  other 
two  may  be  found.    This  may  be  shown  thus : 

Taking  7  as  the  first  term  of  an  increasing  series,  and  5  the  constant 
diflference,  the  series  may  be  written  in  two  forms ;  thus  : 


1st  Term.     Sd  Term. 
(1)  7  13 


(3) 


7  +  (5) 


7  +  (5+5) 


Jith  Term. 

23,  and  so  on. 


7  +  (5  +  5  +  5) 


Observe,  in  (3),  each  term  is  composed  of  the  fi/rst  term  7  plus  as  many 
times  the  constant  difference  5  as  the  number  of  the  term  less  1.  Thus, 
for  example,  the  ninth  term  in  this  series  would  be  7  +  5  x  (9— 1)=47. 

Hence,  from  the  manner  in  which  each  term  is  composed,  we  have  the 
following  forfnulae  or  rules : 


\,  A  =  L-I>x  {N^  1).    Read, 


i  =  ^  +  i>  X  {N-  1).    Read 


.{ 


The  first  term  is  equal  to  the  last 
term.,  minus  the  common  difference 
multiplied  by  the  number  qf  terms 
less  1. 

The  last  term  is  equal  to  the  first 
term,  plus  the  common  difference  mul- 
tiplied by  the  number  of  terms  less  1. 


3.  D  = 


L-A 

N-1' 


Read, 


(334) 


The  common  difference  is  equal  to 
the  last  term,  minus  the  first  term 
divided  by  the  number  of  terms  l«is  L 


PROGRESSION.  207 

jf^  __  ^  r      TJie  number  of  terms  is  equal  to  the 

4.  JV=  — fz —  +  1.  "Read,  -l  last  term,  minus  the  first  term  divided 

■^  [  l^the  common  difference,  plus  1. 

Obsefve,  that  in  a  decreasing  series,  the  first  term  is  the  largest  and 
the  last  term  the  smallest  in  the  series.  Hence,  to  make  the  above  for- 
mulae apply  to  a  decreasing  series,  we  must  place  L  where  A.  is,  and  A. 
where  L  is,  and  read  the  formulae  acccordingly. 

764.  To  show  how  to  find  the  sum  of  a  series  let 

(1.)  4  7        10        13       16       19      be  an  arithmetical  series. 

(2.)        19        16        13        10         7         4     be  the  same  series  reversed. 
(3.)         23  -f  23  +  23  +  23  +  23  -{-  23 =twice  the  sum  of  the  terms. 

Now,  observe,  that  in  (3),  which  is  equal  to  twice  the  sum  of  the 
series,  each  term  is  equal  to  the  first  term  plus  the  last  term  ;  hence, 

The  sum  of  the  terms  of  an  arithmetical 
series  is  equal  to  one-half  of  the  sum  of  the 
first  and  last  term^  multiplied  by  Vie  number 
(f  terms. 


S  =  loi{A  +  L)x  N,    Read,  ■ 


EXAMPLES     FOR     PRACTICE. 

765.  1.  The  first  term  of  an  arithmetical  progression  is  4, 
the  common  difference  2 ;  what  is  the  12th  term  ? 

2.  The  first  round  of  an  upright  ladder  is  12  inches  from 
the  ground,  and  the  nineteenth  246  inches;  how  far  apart  are 
the  rounds  ? 

3.  The  tenth  term  of  an  arithmetical  progression  is  190, 
the  common  difference  20 ;  what  is  the  first  term  ? 

4.  Weston  traveled  14  miles  the  first  day,  increasing 
4  miles  each  day;  how  far  did  he  travel  the  15th  day, 
and  how  many  miles  did  he  travel  in  all  the  first 
12  days  ? 

5.  The  amount  of  $^6^}  for  'i^,years  at  simple  interest  was 
$486 ;  what  was  the  yearly  interest  ? 

^.  The  first  term  of  an  arithmetical  series  of  100  terms 
is  150,  and  the  last  term  1338;  what  is  the  common 
difference  ? 

(335) 


SOS  BUSINUSS    ARITHMETIC. 

7.  What  is  the  sum  of  the  first  1000  numbers  in  their 
natural  order  ? 

8.  A  merchant  bought  16  pieces  of  cloth,  giving  10  cents 
for  the  first  and  $12.10  for  the  last,  the  several  prices  form 
an  arithmetical  series  ;  find  the  cost  of  the  cloth  ? 

9.  A  man  set  out  on  a  journey,  going  6  miles  the  first 
day,  increasing  the  distance  4  miles  each  day.  The  last  day 
he  went  50  miles ;  how  long  and  how  far  did  he  travel  ? 

10.  How  many  less  strokes  are  made  daily  by  a  clock  which 
strikes  the  hours  from  1  to  12,  than  by  one  which  strikes 
from  1  to  24. 


GEOMETRICAL  PROGRESSIOK 

766.  There  are  five  quantities  considered  in  geometrical 
progression,  which  we  denote  by  letters  in  the  same  manner 
as  in  arithmetical  progression ;  thus  : 

\.  A  =  First  Term.  2.  i  =  Last  Term. 

3.  £  =  Constant  Ratio.  ^.  N=  Number  of  Tenns. 

5.  8  =  the  Sum  of  all  the  terms. 

767.  Any  three  of  these  quantities  being  given,  the.  other 
two  may  be  found.     This  may  be  shown  thus : 

Taking  3  as  the  first  term  and  2  as  the  constant  ratio  or  multiplier, 
the  series  may  be  written  in  three  forms' ;  thus : 

l8t  Term.  M  Term.  3d  Term.        4th  Term.  Sth  Term. 

(1.)        8  6  12  24  48 


(2.)        8        3x2        3x(2x2)        3x  (2x2x2)        3x  (2x2x2x2) 
(3.)        3        3x2  3x2*    ^  3x2^  3x2* 

Observe,  in  (3),  each  term  is  composed  of  the  first  term,  8,  multiplied 
by  the  constant  multiplier  2,  raised  to  the  power  indicated  by  the 
number  of  the  term  less  1.  Thus,  for  example,  the  seventh  term  would 
be3x2'-»  =  3x2«  =  192. 

(336) 


PROG  RE8SI0N,  209 

Hence,  from  tlie  manner  in  wliicli  each  term  is  composed,  we  have 
the  following  formulae  or  rules  : 


L 


{ 


The  first  term  is  egual  to  the  last  term,  divided 
1,  A.  =—  _..  Read,  -j   by  the  constant  mvltiplier  raised  to  the  power 

-K""  I  indicated  by  the  number  of  terms  less  1. 


iThe  last  term  is  equal  to  the  first  term,  multi- 
plied by  the  constant  mvltiplier  raised  to  the 
power  indicated  by  the  number  of  terms  less  1. 

r  The  constant  multiplier  is  equal  to  the  root, 
_     _y_n— i/i  -^      ,     I  whose  index  is  indicated  by  th^  numi)er  of  terms 

6.  M—  y  -g^  Iteaa,  J  ^^^^  ^^^  ^  ^^  quotient  of  the  last  term  divided 

I  by  the  first. 

{The  number  of  terms  less  one  is  equal  to  the 
exponent  of  the  power  to  which  the  common 
multiplier  micst  be  raised  to  be  equal  to  tlie 
quotient  qf  the  last  term  divided  by  the  first. 

768.  To  show  how  to  find  the  sum  of  a  geometrical  series, 
we  take  a  series  whose  common  multiplier  is  known  ;  thus : 

>Sf  =  5  +  15  4-  45  -f  135  +  405. 

Multiplying  each  term  in  this  series  by  3,  the  common  multiplier,  we 
will  have  3  times  the  sum. 

(1.)  iS^x  3=5  X 3  +  15x3-1-45x3  +  135x3  +  405x3,  or 
{2.)  Sx3=15     +45       +135     +405        +405x3. 

Subtracting  the  sum  of  the  series  from  this  result  as  expressed  in  (2), 
we  have, 

;Sfx3  =         15  +  45  +  135  +  405+405x3 

8 =  5  +  15  +  45  +  135  +  405 

8x2  =  405x3-5 

Now,  observe,  in  this  remainder  S  -x  2  ib  S  x  (JB  —  1),  and  405  x  3  is 
X  X  li,  and  5  is  A.  Hence,  Sx{It  —  l)  =  Lx  M  —  A.  And  since 
^  —  1  times  the  Sum  is  equal  to  i  x  R  —  A,  we  have, 

(The  mm  of  a  geometrical  seiies  is  equal  to 
the  difference,  between  the  last  term  multiplied 
__  by  the  ratio  and  the  first  term,  divided  by  the 

I  ratio  minus  1. 

(337) 


^10  BUSINESS    ARITHMETIC. 


EXAMPIil^S     FOR     PRACTICE. 

769.  1.  The  first  term  of  a  geometrical  progression  is  3, 
the  ratio  4 ;  what  is  the  8th  term  ? 

2.  The  first  term  of  a  geometrical  progression  is  1,  and  the 
ratio  2  ;  what  is  the  12th  term  ? 

3.  The  extremes  are  4  and  2916,  and  the  ratio  3 ;  what  is 
the  number  of  terms  ? 

4.  The  extremes  of  a  geometrical  progression  are  2 
and  1458,  and  the  ratio  3;  what  is  the  sum  of  all  the 
terms  ? 

5.  The  first  term  is  3,  the  seventeenth  196608 ;  what  is 
the  sum  of  all  the  terms  ? 

6.  A  man  traveled  6  days ;  the  first  day  he  went  5  miles 
and  doubled  "the  distance  each  day;  his  last  day's  ride  was 
160  miles;  how  far  did  he  travel? 

7.  Supposing  an  engine  should  start  at  a  speed  of  3  miles 
an  hour,  and  the  speed  could  be  doubled  each  hour 
until  it  equalled  96  miles,  how  far  would  it  have  moved 
in  all,  and  how  many  hours  would  it  be  in  motion  ? 

8.  The  first  term  of  a  geometrical  progression  is  4,  the  7th 
term  is  2916 ;  what  is  the  ratio  and  the  sum  of  the  series  ? 


ANNUITIES. 

770.  An  Annuity  is  a  fixed  sum  of  money,  payable 
annually,  or  at  the  end  of  any  equal  periods  of  time. 

771.  The  Amount  or  Final  Value  of  annuity  is  the 
sum  of  all  the  payments,  each  payment  being  increased  by 
its  interest  from  the  time  it  is  due  until  the  annuity  ceases. 

772.  The  JPresent  Worth  of  an  annuity  is  such  a  sum 
of  money  as  will  amount,  at  the  given  rate  per  cent,  in  the 
given  time,  to  the  Amount  or  Final  Value  of  the  annuity. 

(338) 


PROGRESSION,  211 

773.  An  Annuity  at    Simple  Interest  forms  an 

arithmetical  progression  whose  common    difference    is    the 
interest  on  the  given  annuity  for  one  interval  of  time. 

Thus  an  annuity  of  $400  for  4  years,  at  7%  simple  interest,  gives 
the  following  progression  : 

1st  Term.  2d  Term.  3d  Term.  ^th  Term. 

(1.)    $400    $400 +  ($28)    $400 +  ($38 +  $28)     $400 +  ($28 +  $28 +  $28),  or 
(2.)    $400         $428  $456 


Observe,  there  is  no  interest  on  the  last  payment ;  hence  it  forms  the 
1st  Term.  The  payment  before  the  last  bears  one  year's  interest,  hence 
forms  the  2d  Term  ;  and  so  on  with  the  other  terms. 

Hence  all  problems  in  annuities  at  simple  interest  are  solved  by 
arithmetical  progression. 

774.  An  Annuity  at  Compound  Interest  forms  a 

geometrical  progression  whose  common  multiplier  is  repre- 
sented by  the  amount  of  $1  for  one  interval  of  time. 

Thus  an  annuity  of  $300  for  4  years,  at  6%  compound  interest,  gives 
the  following  progression  : 

1st  Term.    2d  Term.  3d  Term.  4th  Term. 

$300  X  1.06        $300  X  1.06  x  1.06        $300  x  1.06  x  1.06  x  1.06. 


Observe  carefully  the  following  : 

(1.)  The  last  payment  bears  no  interest,  and  hence  forms  the  1st  Term 
of  the  progression. 

(2.)  The  payment  before  the  last,  when  not  paid  until  the  annuity 
ceases,  bears  interest  for  one  year  ;  hence  its  amount  is  $300  x  1.06  and 
forms  the  2d  Term. 

(3.)  The  second  payment  before  the  last,  bears  interest  when  the 
annuity  ceases, for  two  years ;  hence  its  amount  at  compound  interest  is 
$300  X  1.06,  the  amount  for  one  year,  multiplied  by  1.06,  equal  $300  x 
1.06  X  1.06,  and  forms  the  od  Term,  and  so  on  with  other  terms. 

Hence  all  problems  in  annuities  at  compound  interest  are  solved  by 
geometrical  progression. 

(339) 


212  BUSINESS    ARITHMETIC, 


SXAMPIiES     FOR     PRACTICE. 

775.  1.  What  is  the  amount  of  an  annuity  of  $200  for 
6  years  at  11%  simple  interest  ? 

2.  A  father  deposits  $150  annually  for  the  benefit  of  his 
son,  beginning  with  his  12th  birthday;  what  will  be  the 
amount  of  the  annuity  on  his  21st  birthday,  allowing  simple 
interest  at  6%  ? 

3.  What  is  the  present  worth  of  an  annuity  of  $600  for 
5  years  at  8^,  simple  interest  ? 

4.  What  is  the  amount  of  an  annuity  of  $400  for  4  years  at 
7%,  compound  interest  ? 

5.  What  is  the  present  worth  of  an  annuity  of  $100  for 
G  years  at  6%,  compound  interest  ? 

6.  What  is  the  present  worth  of  an  annuity  of  $700  at  S%, 
simple  interest,  for  10  years  ? 

7.  What  is  the  amount  of  an  annuity  of  $500  at  '7%,  com- 
pound interest,  for  12  years? 

8.  What  is  the  present  worth  of  an  annuity  of  $350  for 
9  years  at  6^,  compound  interest  ? 

This  example  and  the  four  following  should  be  solved  by  applying 
the  formulae  for  geometrical  progression  on  page  337. 

10.  At  what  rate  ^  will  $1000  amount  to  $1500.73  in  6  years, 
compound  interest  ? 

11.  The  amount  of  a  certain  sum  of  money  for  12  years, 
at  7^  compound  interest,  was  $1126.096;  what  was  the 
original  sum  ? 

12.  What  sum  at  compound  interest  8  years,  at  7^  will 
amount  to  $4295.465  ? 

13.  In  how  many  years  will  $20  amount  to  $23.82032,  at 
6^  compound  interest? 

(340) 


saf/yr'^ggyf^^. 


.5^ 


[mensuration] 


-^^L^2jj±^ 


^l^wp^ 


GENERAL    DEFINITIONS. 

776.  A  Line  is  that  whicli  has  only  length. 

777.  A  Straight  Line  is  a  line  which  has  the  same  direction  at 
every  point. 

778.  A  Curved  Line  is  a  line  which  changes  its  direction  at 
every  point. 

779.  Parallel  Lines  are  lines  which  have  the  same  direction. 

780.  An  Angle  is  the  opening  between  two  lines  which  meet  in  a 
common  point,  called  the  vertex. 


Angles  are  of  three  kinds,  thus  : 

(1)                          (2) 

(3)^ 

(4) 

u 

> 

1                               1 
4            ..c 

)i ^ 

OUme  Angle. 

'A 

HORIZ 

TwoRi 

ONTAl..      '                  A 

ght  Angles.          On 

c 
e  Eight  Angle. 

Acu 

r —    i 

te  Angle. 

781.  When  a  line  meets  another  line,  making,  as  shown  in  (1),  two 
equal  angles,  each  angle  is  a  Right  Angle,  and  the  lines  are  said  to 
be  perpendicular  to  each  other. 

782.  An  Obtuse  Angle,  as  shown  in  (3),  is  greater  than  a  right 
angle,  and  an  Acute  Angle,  as  shown  in  (4),  is  less  than  a  right  angle. 

Angles  are  read  by  using  letters,  the  letter  at  the  vertex  being  always 
read  in  the  middle.    Thus,  in  (2),  we  read,  the  angle  BAG  or  GAB. 

783.  A  Plane  is  a  surface  such  that  if  any  two  points  in  it  be 
joined  by  a  straight  line,  every  point  of  that  line  will  be  in  the  surface. 

(341) 


214 


BUSINESS     ARITHMETIC. 


784.  A  Plane  Figure  is  a  plane  bounded  either  by  straight  or 
curved  lines,  or  by  one  curved  line. 

185.   A  Polygon  is  a  plane  figure  bounded  by  straight  lines.    It 
is  named  by  the  number  of  sides  in  its  boundary  ;  thus : 


Trigon. 


Tetragon. 


Pentagon.       Hexagon,  and  so  on. 


Observe,  that  a  regular  polygon  is  one  that  has  all  its  sides  and  all  its 
angles  equal,  and  that  the  Base  of  a  polygon  is  the  side  on  which  it  stands. 

^S(y,  A  Trigon  is  a  <Are<?-sided  polygon.    It  is  usually  called  a 
Triangle  on  account  of  having  three  angles. 

Triangles  are  of  three  kinds,  thus  : 


G) 


0 

A 


A  C 

Right-angled  Triangle. 


A        o       c 
Acute-angled  Triangle. 


Obtuse-angled  Triangle. 


Observe,  a  right-angled  triangle  has  one  Hght  angle,  an  acute-angled 
triangle  has  three  acute  angles^  and  an  obtuse-angled  triangle  has  one 
obtuse  angle. 

Observe,  also,  as  shown  in  (2)  and  (3),  that  the  Altitude  of  a  triangle  is 
the  perpendicular  distance  from  one  of  its  angles  to  the  side  opposite. 

787.  An  Equilateral  Triangle  is  a  triangle  whose  three  sides 
are  equal. 

788.  An  Isosceles  Triangle  has  two  of  its  sides  equal. 

789.  A  Scalene  Triangle  has  all  of  its  sides  unequal. 

790.  A  Tetragon  is  a  four-sided  polygon.  It  is  usually  called  a 
Quadrilateral. 

(343) 


MENSURATION, 


%u 


Quadrilaterals  are  of  three  kinds,  thus  : 

(1)  ^- 


Parallelogram. 


Trapezoid. 


Trapezium. 


Observe,  that  a  Parallelogram  has  its  opposite  sides  parallel,  that  a 
Trapezoid  has  only  two  sides  parallel,  and  that  a  Trapezium  has  no  sides 
parallel. 

Observe,  also,  that  the  Diagonal  of  a  quadrilateral,  as  shown  in  (1),  (2) 
and  (3),  is  a  line  joining  any  two  opposite  angles. 

"JOl.  A  Parallelogram  is  a  quadrilateral  which  has  its  opposite 
sides  parallel.     Parallelograms  are  of  four  kinds,  thus : 


(1) 


(2) 


Beciangle. 


BhomM,d. 


Eh&inMs. 


Observe,  that  a  Square  has  all  its  sides  equal  and  all  its  angles  right 
angles,  that  a  Rectangle  has  its  opposite  sides  equal  and  nU  its  angles 
right  angles,  that  a  Rhomboid  has  its  opposite  sides  equal  and  its  angles 
acute  and  obtuse,  and  that  a  Rhombus  has  all  its  sides  equal  and  its 
angles  nctite  and  obtuse. 

Observe,  also,  that  the  Altitude  of  a  parallelogram,  as  shown  in  (3)  and 
(4),  is  the  perpendicular  distance  between  two  opposite  sides. 

793.  A  Circle  is  a  plane  bounded  by  a  curved  line,  called  the 
drcMmference,  every  point  of  which  is  equally  distant  from  a  point 
within,  called  the  centre  ;  thus : 

793.  The  Diayneter  of  a  circle  is  any  straight 
line,  as  CD,  passing  through  its  centre  and  terminating 
at  both  ends  in  the  circumference. 

794.  The  HadiuH  of  a  circle  is  any  straight  line, 
as  AB,  extending  from  the  centre  to  the  circumference. 

(343) 


216 


BUSINESS    ARITHMETIC. 


795.  The  Perimeter  of  a  polygon  is  the  sum  of  all  the  lines  which 
form  its  boundary,  and  of  a  circle  the  circumference. 

796.  The  Area  of  any  plane  figure  is  the  surface    contained 
within  its  boundaries  or  boundary. 

797.  Mensuration  treats  of  the  method  of  finding  the  lengths 
of  lines,  the  area  of  surfaces,  and  volumes  of  solids. 

SOLUTION  OF  PROBLEMS. 


(1) 

8  units  long. 

T     ' 

.,.^. 

-' 

^g 

^^%^=bri 

- 

798.  The  solutions  of  problems  in  mensuration  cannot  be  demon- 
Btrated  except  by  geometry,  but  the  general  principle  which  underlies 
these  solutions  may  be  stated ;  thus, 

Tlie  contents  of  any  given  surface  or  solid  that  can  he  meamrcd  can  he 
shown  to  he  equal  to  the  contents  of  a  rectangular  surface  or  solid,  whose 
dimensions  are  equal  to  certain  known  dimensions  of  the  given  surface 
or  solid,  thus : 

1.  Observe,  that  the  number  of  small  squares 
in  (1)  is  equal  to  the  product  of  the  numbers 
denoting  the  length  and  breadth.  Thus,  8x5= 
40  small  squares. 

2.  Observe,  in  (2),  that  the  plane  bounded  by 
the  lines  FB,  BC,  CE,  and  EF,  is  rectangular 
and  equal  to  the  given  parallelogram  ABCD, 
because  we  have  added  to  the  right  as  much 
surface  as  we  have  taken  off  at  the  left. 

Hence    the    contents    of    the    parallelogram 
ABCD  is  found  by  taking  the  product  of  the 
number  of  units  in  the    altitude    CE   or  BF, 
and  in  the  side  BC. 

3.  Observe,  again,  the  diagonal  BD  divides  the  parallelogram  into 
two  equal  triangles,  and  hence  the  area  of  the  triangle  ABD  is  one-half 
the  area  of  the  parallelogram,  and  is  therefore  found  by  taking  one-half 
of  the  product  of  the  number  of  units  in  the  base  AD  and  in  the 
altitude  BF  or  CE. 

In  view  of  the  fact  that  the  solutions  in  mensuration  depend  upon 
geometry,  no  explanations  are  given.     The  rule,  in  each  case,  must  bo 


strictly  followed. 


(344) 


MENSURATIOIV.  217 

PROBLEMS  ON  TRIANGLES. 

799.  Prob.  I. —  When  the  base  and  altitude  of  a  triangle  are  given, 
to  find  the  area :  Divide  the  product  of  the  base  and  ALTrruDE  hy  2, 

Find  the  area  of  a  triangle 

1.  Whose  base  is  14  ft.  and  altitude  7  ft.  8  in. 

2.  Whose  base  is  3  rd.  and  altitude  2  rd.  7  ft. 

3.  Whose  base  is  21  chains  and  altitude  16  chains. 

4.  What  is  the  area  of  a  triangular  park  whose  base  is  16.76  chains 
and  altitude  13.4  chains  ? 

5.  How  many  square  feet  of  lumber  will  be  required  to  board  up  the 
gable-ends  of  a  house  30  feet  wide,  having  the  ridge  of  the  roof  17  feet 
higher  than  the  foot  of  the  rafters  ? 

6.  How  many  stones,  each  2  ft.  6  in.  by  1  ft.  9  in.  will  be  used  in 
paving  a  triangular  court  whose  base  is  1 50  feet  and  altitude  126  feet, 
and  what  will  be  the  expense  at  $.35  a  square  yard  ? 

800.  Prob.  IL — TFA^ti  the  area  and  one  dimension  are  gimn,  to 
find  the  other  dimension :  Double  the  area  and  divide  by  the  given 
dimension. 

Find  the  altitude  of  a  triangle 

1.  Whose  area  is  75  square  feet  and  base  15  feet. 

2.  Whose  area  is  264  square  rods  and  base  24  rods. 

3.  Whose  base  is  6  ft.  1  in.  and  area  50  sq.  ft.  100  sq.  in. 
Find  the  base  of  a  triangle 

4.  Whose  area  is  3  A.  108  P.  and  altitude  28  rd. 

5.  Whose  altitude  is  2  yd.  2  ft.  and  area  8  sq.  yd. 

6.  Whose  area  is  2  sq.  rd.  19  sq.  yd.  2  sq.  ft.  36  sq.  in.  and  altitude 

1  rd.  1  ft.  6  in. 

7.  For  the  gable  of  a  church  75  feet  wide  it  required  250  stones,  each 

2  ft.  long  and  1  ft.  6  in.  wide  ;  what  is  the  perpendicular  distance  from 
the  ridge  of  the  roof  to  the  foot  of  the  rafters  ? 

801.  Prob.  IH. —  When  the  three  sides  of  a  triangle  are  given,  to 
find  the  area :  From  half  the  sum  of  the  three  sides  subtract  each  side 
separately.  Multiply  the  half  sum  and  the  three  remainders  together ; 
the  square  root  of  the  'product  is  the  area. 

1.  Find  the  area  of  a  triangle  whose  sides  are  15,  20,  25  feet. 

2.  What  is  the  area  of  an  isosceles  triangle  whose  base  is  50  in.  and 
each  of  its  equal  sides  35  inches  ? 

(345) 


218  BUSINESS    ARITHMETIC 

3.  How  many  acres  in  a  triangular  field  whose  sides  measure  16,  20, 
30  rods? 

4.  What  is  the  area  of  an  equilateral  triangle  whose  sides  each 
measure  40  feet  ? 

5.  A  piece  of  land  in  the  form  of  an  equilateral  triangle  requires 
158  rods  of  fence  to  enclose  it ;  how  many  acres  are  there,  and  what  is 
the  cost  at  $40  per  acre  ? 

803,  Prob.  TV.—  W7ien  tJie  base  and  perpendicular  are  given  in  a 
rigid-angled  triangle,  to  find  the  other  side  :  Extract  the  square  root  of  the 
sum  of  the  squares  of  the  base  and  perpendicular. 

The  reason  of  this  rule  and  the  one  in  Prob.  V  will  be  seen  by 
examining  the  diagram  in  the  margin. 

Observe,  that  the  square,  on  the  side  AB  opposite 
the  right  an|?le,  contains  as  many  small  squares  as 
the  sum  of  the  small  squares  in  the  squares  on  the 
base  AC  and  the  perpendicular  BC.  This  is  shown 
by  geometry  to  be  true  of  all  right-angled  triangles. 
Hence,  by  extracting  the  square  root  of  the  sum 
of  the  squares  of  the  base  and  perpendicular  of  a 
right-angled  triangle,  we  have  the  length  of  the 
side  opposite  the  right  angle. 

The  side  opposite  the  right  angle  is  called  the  JSypothentlSe. 

Find  the  hypothenuse  of  a  right-angled  triangle 

1.  Whose  base  is  40  ft.  and  perpendicular  16  ft. 

2.  Whose  base  is  15  ft.  and  perpendicular  36  ft. 

3.  A  tree  104  ft.  high  stands  upon  the  bank  of  a  stream  76  feet  wide ; 
what  is  the  distance  of  a  man  upon  the  opposite  bank  from  a  raven  .upon 
the  top  of  the  tree  ?  '         O^  I 

4.  A  and  B  start  from  one  comer  of  a  field  a  mile  square,  traveling 
at  the  same  rate ;  A  follows  the  fence  around  the  field,  and  B  proceeds 
directly  across  to  the  opposite  corner ;  when  B  reaches  the  corner,  how 
far  will  he  be  from  A  ? 

5.  What  is  the  length  of  the  shortest  rope  by  which  a  horse  may  be 
tied  to  a  post  in  the  middle  of  a  field  20  rods  square,  and  yet  be  allowed 
to  graze  upon  every  part  of  it  ? 

(346) 


C 


MENSURATION,  ^t^ 

803.  Peob.  Y.—  When  the  bme  or  perpendicular  is  to  he  found  : 
Extract  the  square  root  of  the  difference  between  tlie  aqvxire  of  the  hypother 

nuse  and  the  square  of  the  given  side. 

Find  tlie  base  of  a  right-angled  triangle 

1.  Whose  hypothenuse  is  40  ft.  and  perpendicular  15  feet. 

2.  Whose  perpendicular  is  20  feet  and  hypothenuse  45  feet. 

3.  Bunker  Hill  monument  is  220  feet  high  ;  a  man  360  feet  from  the 
base  shot  a  bird  hovering  above  the  top  ;  the  man  was  423  feet  from  the 
bird ;  how  far  was  the  bird  from  the  top  of  the  monument  ? 

4.  A  ladder  35  feet  long  reaches  from  the  middle  of  the  street  to  a 
window  28  feet  high  ;  how  wide  is  the  street  ? 

5.  The  lower  ends  of  two  opposite  rafters  are  48  feet  apart  and  the 
length  of  eacli  rafter  is  30  feet ;  what  is  the  elevation  of  the  ridge  above 
the  eaves  ? 

PROBLEMS  ON  QUADRILATERALS. 

804.  Prob.  VI. — 2''o  find  the  area  of  a  pa/rallelogram  :  Multiply 
tlie  base  by  th^  altitude. 

1.  Find  the  area  of  a  parallelogram  whose  base  is  3  ft.  9  in.  and  alti- 
tude 7  ft.  8  in.  ;  whose  altitude  is  2  yd.  5  in.  and  base  3  yd.  6  in. 

2.  How  many  acres  in  a  piece  of  land  in  the  form  of  a  parallelogram 
whose  base  is  9.86  ch.  and  altitude  7.5  ch? 

3.  How  many  square  feet  in  the  roof  of  a  building  85  ft.  long,  and 
whose  rafters  are  each  16  ft.  0  in.  long  ? 

4.  The  base  of  a  rhombus  is  9  ft.  8  in.  and  its  altitude  3  ft.  ;  how 
many  square  feet  in  its  surface  ? 

805.  Prob.  VII. — To  find  the  area  of  a  trapezoid:  Multiply 
one-half  of  the  sum  of  the  parallel  sides  by  the  altitude. 

Find  the  area  of  a  trapezoid 

1.  Whose  parallel  sides  are  15  and  25  feet  and  altitude  11  feet. 

2.  Whose  parallel  sides  are  8  and  11  inches  and  altitude  6  inches. 

3.  How  many  square  feet  in  a  board  1  ft.  4  in.  wide,  one  side  of 
which  is  32  ft.  long  and  the  other  side  34  feet  long  ? 

4.  One  side  of  a  field  measures  47  rods,  the  side  opposite  and  parallel 
to  it  measures  39  rods,  and  the  distance  between  the  two  sides  is 
15  rods ;  how  much  is  it  worth  at  $40  per  acre  ? 

(347) 


220  BUSINESS    ARITHMETIC. 

806.  Pbob.  VIII. — To  find  the  area  of  a  trapezium :  Multiply 
the  diagonal  by  half  the  sum  of  the  perpendicidara  to  it  from  the 
opposite  angles. 

Refer  to  diagram  (3)  in  (790)  and  find  tlie  area  of  a  trapezium 

1.  Whose  diagonal  is  45  in.  and  perpendiculars  to  this  diagonal 
11  inches  and  9  inches. 

2.  Whose  diagonal  is  16  feet  and  perpendiculara  to  this  diagonal 
7  feet  and  6  feet. 

8.  Whose  diagonal  is  37  ft.  6  in.  and  perpendiculars  to  this  diagonal 
7  ft.  4  in.  and  8  ft.  8  in. 

4.  How  many  acres  in  a  field  in  the  form  of  a  trapezium  whose 
diagonal  is  \  mi.  and  the  perpendiculars  to  this  diagonal  5  ch. 
and  6  ch.  ? 

807.  Prob.  IK.  — To  find  the  diameter  of  a  evrde:  Dimde  the 
circumference  by  3.14I6. 

To  find  the  circumference :  Multiply  the  diameter  by  3.14I6. 

1.  Find  the  diameter  of  a  circle  whose  circumference  is  94.248  inches ; 
whose  circumference  is  78.54  feet. 

2.  Find  the  circumference  of  a  circle  whose  diameter  is  14  inches  ; 
whose  radius  is  9  inches. 

3.  What  will  it  cost  to  fence  a  circular  park  3  rods  in  diameter, 
at  $4.80  per  rod. 

4.  How  many  miles  does  the  earth  pass  over  in  its  revolution  around 
the  sun,  its  distance  from  the  sun  being  95,000,000  miles? 

808.  Prob,  X.—To  find  tJie  area  of  a  circle :  Multiply  \  of  its 
diameter  by  the  circumference  ;  or.  Multiply  the  square  of  its  diameter 
by  .7854. 

1.  What  is  the  area  of  a  circle  whose  diameter  is  20  feet  ?  Whose 
diameter  is  42  inches?    Whose  circumference  is  157.08  feet  ? 

2.  What  is  the  area  of  the  largest  circular  plot  that  can  he  cut 
from  a  field  135  feet  square?  How  much  must  be  cut  off  at  the  comers 
in  making  this  plot  ?  How  much  less  will  it  cost  to  fence  this  than  the 
square,  at  $2.50  a  rod  ? 

3.  The  distance  around  a  circular  park  is  1^  miles.  How  many  acrea 
does  it  contain  ? 

(348) 


MENSURATION, 


221 


809.  PPvOb.  XL.— To  find  the  diameter  when  the  area  of  a 
circle  is  given  :  Extract  the  square  root  of  the  quotient  of  tJie  area  divided 
by  .7854. 

Ohserce,  that  when  the  diameter  is  found,  the  circumference  can  be 
found  by  multiplying  the  diameter  by  3. 1416  (SOT). 

1.  What  is  the  diameter  of  a  circle  whose  area  is  50.2656  sq.  ft.  ? 

2.  What  is  the  circumference  of  a  circle  whose  area  is  153.9384 
square  feet  ? 

3.  The  area  of  a  circular  lot  is  19.635  square  rods  ;  what  is  its 
diameter  ? 

4.  The  area  of  a  circle  is  113.0976  sq.  in.  ;  what  is  its  circumference  ? 

5.  How  many  rods  of  fence  will  be  required  to  inclose  a  circle  whose 
area  is  3142%  square  rode  ? 

C.  What  is  the  radius  of  a  circle  whose  area  is  804.2496  sq.  in.? 


PROBLEMS  ON  SOLIDS  OR  VOLUMES. 

810,  A   Solid    or    Volume    has    three    dimensions :    length, 
breadth,  and  thickness. 

The  boundaries  of  a  solid  are  planes.     They  are  called  faoes,  and 
their  intersections  edges. 

811.  A  Prism  is  a  solid  or  volume  having  two  of  its  faces  equal 
and  parallel  polygons,  and  its  other  faces  parallelograms, 

Observe,  a  prism  is  named  by  the  number  of  sides  in  its  equal  and 
parallel  faces  or  bases  ;  thus : 

(1)  (2)  (3) 


Quadrangular 
Prism. 


Pentagonal 
Prism. 


Observe,  a  Prism  whose  parallel  faces  or  bases  are  parallelograms,  as 
shown  in  (2),  is  called  a  Paralleloxripedon, 

Observe,  also,  that  the  Altitude  of   a  prism  is  the  perpendicular 
distance  between  its  bases. 

(349) 


223 


BUSINESS    ARITHMETIC, 


81^1  A  Cylinder,  as  shown  in  (1),  is  a  round  solid  or  volume 
having  two  equal  and  parallel  circles  as  its  hoses, 

1.  Observe,  that  the  altitude  of  a  cylinder 
is  the  perpendicular  distance  between  the 
two  circles  forming  its  bases. 

2.  Observe,  also,  that  a  cylinder  is  con- 
ceived to  be  generated  by  revolving  a 
rectangle  about  one  of  its  sides. 

813.  A  Sphere,  as  shown  in  (2),  is  a  solid  or  volume  bounded  by 
a  curved  surface,  such  that  all  points  in  it  are  equally  distant  from  a 
point  within,  called  the  centre. 

814.  The  Diameter  of  a  sphere  is  a  line,  as  CD  in  (2),  passing 
through  its  centre  and  terminating  at  both  ends  in  the  surface. 

815.  The  JRadius  of  a  sphere  is  a  line  drawn  from  the  centre  to 
any  point  in  the  surface. 

816.  A  Pyramid,  as  shown  in  (1),  is  a  solid  or  volume  having 
as  its  base  any  polygon,  and  as  its  other  faces  triangles,  which  meet  in  a 
common  point  called  the  vertex. 


Pyramid. 


J?rustum, 


Cone. 


Frustum. 


817.  A  Cone,  as  shown  in  (3),  is  a  solid  or  volume  whose  base  is  a 
circle  and  whose  convex  surface  tapers  uniformly  to  a  point,  called  the 
fiertex. 

1.  Observe,  that  the  AltUude  of  a  pyramid  or  cone  is  the  perpendictilar 
distance  between  the  vertex  and  the  base. 

2.  Observe,  also,  that  the  Slant  Height  of  a  pyramid  is  the  perpen- 
dicular distance  between  the  vertex  and  one  of  the  sides  of  the  base ; 
and  of  a  cone  the  distance  between  the  vertex  and  the  circumference  of 
the  base. 

(350) 


MENSURATION. 


818.  A  Frustum  of  a  pyramid  or  cone,  as  shown  in  (3)  and  (4), 
is  the  j)art  which  remains  after  cutting  off  the  top  by  a  plane  parallel  to 
the  base. 

819.  Prob.  XII. — To  find  the  convex  surface  of  a  prism  or  cylinder  : 

Multiply  the  perimeter  of  the  base  by  tJie  altitude. 
To  find  the  entire  surface  add  the  area  of  the  bases. 

The  reason  of  this  rule  may  be  shown  thus : 

(1)  (3)  (3) 


Observe,  that  if  the  three  faces  of  the  prism  in  (1)  are  marked  out 
side  by  side,  as  shown  in  (2),  we  have  a  rectangle  which  is  equal  to  the 
convex  surface  in  the  prism. 

Observe,  also,  that  the  surface  of  the  cylinder  in  (3)  may  he 
conceived  as  spread  out,  as  shown  in  (2) ;  hence  the  reason  of 
the  rule. 

Find  the  area  of  the  convex  surface 

1.  Of  a  prism  whose  altitude  is  8  feet,  and  its  base  a  triangle,  the 
sides  of  whose  base  measures  4  ft. ,  3  ft. ,  3  ft.  6  in. 

2.  Of  a  cylinder  whose  altitude  is  4  ft.  9  in.  and  the  circumference  o 
its  base  7  ft.  8  in. 

3.  Of  a  prism  whose  altitude  is  9  inches,  and  its  base  a  hexagon, 
each  side  of  which  is  2|  inches. 

4.  Find  the  entire  surface  of  a  parallelopipedon  9  ft.  long,  5  ft.  6  in. 
wide,  and  3  ft.  high. 

5.  Find  the  entire  surface  of  a  cylinder  9  ft.  high,  the  diameter  of 
whose  base  is  8  ft. 

820.     Prob.  XllL—To  find  the  volume  of  any  prism  or  cylinder  : 

Multiply  the  area  of  the  base  by  the  altitude. 

1.  Find  the  volume  of  a  triangular  prism  whose  altitude  is  15  ft. 
and  the  sides  of  the  base  each  4  ft. 

2.  What  is  the  volume  of  a  triangular  prism  whose  altitude 
is  28  ft.,  and  the  sides  of  its  base  6  ft.,  7  ft.,  5  ft.  respectively. 


(351) 


224  BUSINESS     ARITHMETIC, 

3.  What  is  the  volume  of  a  parallelopipedon  15  ft.  long,  12  ft.  high, 
10  ft.  wide  ? 

4.  Find  the  contents  of  a  cylinder  whose  altitude  is  19  ft.  and  the 
diameter  of  its  base  4  ft. 

5.  What  is  the  value  of  a  piece  of  timber  15  in.  square  and  50  feet 
long,  at  40  cents  a  cubic  foot  ? 

6.  A  log  is  30  ft.  long  and  its  diameter  is  16  in. ;  how  many  cubic 
feet  does  it  contain  ? 

831.     Prob.  XIV.—  To  find  the  convex  mirface  of  a  pyramid  or  cone : 
Mtdtiply  the  perimeter  of  the  base  hy  one-half  the  slant  height. 
To  find  the  entire  surface,  add  the  area  of  the  hose. 

Find  the  convex  surface  of  a  cone 

1.  Whose    base  is  19    in.  in  circumference,  and  the  slant  height 
12  inches. 

2.  Whose  slant  height  is  15  feet,  and  the  diameter  of  the  base 
10  feet. 

Find  the  convex  surface  of  a  pyramid 

3.  Whose  base  is  3  ft.  6  in.  square,  and  the  slant  height  5  ft. 

4.  Whose  slant  height  is  19  ft.,  and  the  base  a  triangle  whose  sides 
are  12, 14,  8  ft. 

Find  the  entire  surface  of  a  pyramid 

5.  Whose  slant  height  is  45  feet,  and  the  base  a  rectangle  7  ft.  long 
and  8  ft.  wide. 

6.  Whose  slant  height  is  56  in.,  and  its  base  a  triangle  each  of  whose 
sides  is  6  in. 

Find  the  entire  surface  of  a  cone 

7.  Wliose  slant  height  is  42  feet,  and  the  circumference  of  the 
base  31.41G  ft. 

8.  Whose  slant  height  is   75  in.,  and  the  diameter  of   the  base 
5  inches. 

823.     Prob.  XV.—To  find  the  volume  of  a  pyramid  or  cone  : 
Multiply  the  area  of  the  base  hy  one-third  the  altitude. 

Find  the  volume  of  a  cone 

1.  Whose  altitude  is  24  ft.,  and  the  circumference  of   the   base 
C.2832  feet. 

(352) 


MENSURATION.  225 

2.  Whose  altitude  is  12  ft.,  and  the  diameter  of  the  base  4  ft. 

Find  the  volume  of  a  pyramid 

8.  Whose  altitude  is  15  feet,  and  its  base  4  feet  square. 

4.  Whose  altitude  is  18  in.,  and  the  base  a  triangle  8  in.  on 
each  side. 

5.  Whose  altitude  is  45  ft.,  and  its  base  a  rectangle  15  feet 
by  16  feet. 

823.  Prob.  XYl.—Tofind  tJie  convex  surface  of  a  frustum  of  a 
pyramid  or  cone  :  Multiply  the  sum  of  the  perimeters  or  circumf&ren/xa 
by  one-half  the  slant  height. 

To  find  the  entire  surface,  add  the  area  of  both  the  bases. 

1.  What  is  the  convex  surface  of  a  frustum  of  a  triangular  pyramid 
whose  slant  height  is  6  feet,  each  side  of  the  greater  base  3  feet,  and  of 
tlie  less  base  2  feet  ? 

2.  What  is  the  convex  surface  of  a  frustum  of  a  cone  whose  slant 
height  is  9  inches,  and  the  circumference  of  the  lower  base  17  inches, 
and  of  the  upper  base  13  inches  ? 

3.  Find  the  entire  surface  of  a  frustum  of  a  pyramid  whose  slant 
height  is  14  feet,  and  its  bases  triangles,  each  side  of  the  larger  base 
being  8  feet,  and  of  the  smaller  base  6  feet. 

4.  Find  the  entire  surface  of  a  frustum  of  a  cone  whose  slant  height 
is  27  feet,  the  circumference  of  the  greater  base  being  37.6993  feet,  and 
of  the  less  base  31.416  feet. 

834.  Prob.  XVII. — To  find  the  volume  of  a  frustum  of  a  pyramid 
or  cone :  To  the  sum  of  the  areas  of  both  bases  add  the  square 
root  of  their  product  and  multiply  the  re»ult  by  one-third  of  the 
altitude. 

1.  Find  the  volume  of  a  frustum  of  a  square  pyramid  whose 
altitude  is  6  feet,  and  each  side  of  the  lower  base  16  feet,  and  of  the 
upper  base  12  feet. 

2.  How  many  cubic  feet  in  a  frustum  of  a  cone  whose  altitude  is 
9  feet,  the  diameters  of  its  bases  8  feet  and  6  feet. 

3.  How  many  cubic  feet  in  a  section  of  a  tree-trunk  20  feet  long, 
the  diameter  of  the  lower  base  being  18  inches,  and  of  the  upper  base 
12  inches  ? 

4.  One  of  the  big  trees  of  California  is  32  feet  in  diameter  at  the 
foot  of  the  tree  ;  how  many  cubic  feet  in  a  section  of  this  tree  90  feet 
high,  the  upper  base  being  20  feet  in  diameter  ? 

(353) 


226  BUSTNUiSS     ARITHMETIC, 

5.  A  granite  rock,  whose  form  is  a  frustum  of  a  triangular 
pyramid,  is  40  feet  liigli,  the  sides  of  the  lower  base  being  30  feet 
each,  and  of  the  upper  base  16  feet  each.  How  many  cubic  feet 
in  the  rock. 

825.  Prob.  XVIII.— 21?  find  tlie  surface  of  a  sphere :  Multiply 
the  diameter  hy  the  circumference  of  a  great  circle  of  the  given  sphere. 

1.  Find  the  surface  of  a  sphere  whose  diameter  is  8  feet. 

2.  What  is  the  surface  of  a  globe  9  in.  in  diameter  ? 

3.  How  many  square  feet  in  the  surface  of  a  sphere  45  feet  in 
diameter  ? 

4.  What  is  the  surface  of  a  globe  whose  radius  is  1  ft.  6  in.  ? 

5.  How  many  square  inches  in  the  surface  of  a  globe  5  inches  in 
diameter  ? 

836.  Prob.  XIX. — To  find  the  volume  of  a  sphere :  Multiply  the 
surface  by  one-sixth  of  the  diameter. 

1.  Find  the  volume  of  a  sphere  whose  diameter  is  20  inches. 

2.  How  many  cubic  yards  in  a  sphere  whose  diameter  is  3  yards? 

3.  Find  the  solid  contents  of  a  globe  2  ft.  6  in.  in  diameter. 

4.  How  many  cubic  feet  in  a  globe  9  inches  in  diameter  ? 

5.  Find  the  volume  of  a  globe  whose  radius  is  4  inches. 

827.  Prob.  XX.— To  find  the  capacity  of  casks  in  gallons: 
Multiply  the  number  of  inches  in  the  length  by  the  square  of  tJie 
number  of  inches  in  the  mean  diameter,  and  this  product  by  .0034.     ' 

Observe,  that  the  mean  diameter  is  found  (nearly)  by  adding  to  tho 
head  diameter   |,  or  if  the  staves  are  but  slightly  curved,  f  of  the 
diflerence  between  the  head  and  bung  diameters. 
.  The  process  of  finding  the  capacity  of  casks  is  called  Gauging. 

1.  How  many  gallons  in  a  cask  whose  head  diameter  is  20,  bung 
diameter  26  inches,  and  its  length  30  inches  ? 

2.  How  many    gallons  will  a  cask  hold  whose  head  diameter  is 

21  inches,  bung  diameter  30  inches,  and  length  43  inches? 

3.  What  is  the  volume  of  a  cask  whose  diameters  are  18  and 
24  inches  respectively,  and  the  length  32  inches  ? 

4.  A  cask  slightly  curved  is  40  inches  long,  its  head  diameter  being 

22  inches,  and  its  bung  diameter  27  inches;  how  many  gallons  will 
it  hold  ? 

(354) 


-^^^^3^^:^^ 


828.  In  using  this  set  of  review  and  test  examples,  the 
following  suggestions  should  be  carefully  regarded : 

1.  The  examples  cover  all  the  important  subjects  in  arith- 
metic, and  are  designed  as  a  test  of  the  pupil's  strength  in 
solving  difficult  problems  and  of  his  knowledge  of  principles 
and  processes. 

2.  The  teacher  should  require  the  pupil  to  master  the 
thought  expressed  in  each  example  before  attempting  a 
solution. 

To  do  this  he  must  notice  carefully  the  meaning  of  each 
sentence,  and  especially  the  technical  terms  peculiar  to  arith- 
metic; he  must  also  locate  definitely  the  business  relations 
involved. 

3.  When  the  solutions  are  given  in  class,  the  teacher  should 
require  the  pupils  to  state  clearly : 

(a).   What  is  given  and  what  is  required  in  each  example. 

(b).  The  relations  of  the  given  quantities  from  which  ivhat 
is  required  can  he  found. 

{c).  The  steps  that  must  be  taken  in  their  order,  and  the 
processes  that  must  be  used  to  obtain  the  required  result. 

In  making  these  three  statements,  no  set  form  should  be 
used ;  each  pupil  should  be  left  free  to  pursue  his  own  course 
and  give  his  own  solution.  Clearness,  accuracy,  and  brevity 
should  be  the  only  conditions  imposed. 

When  the  work  is  written  on  a  slate,  paper,  or  blackboard, 
neatness  and  a  logical  order  in  arranging  the  steps  in  the 
solution  should  be  invariably  required. 

(355) 


REVIEW    EXAMPLES. 

1.  A  gentleman  held  a  note  for  $1643.20,  payable  in  8  mo., 
without  interest.  He  discounted  the  note  at  %%  for  ready 
cash,  and  invested  the  proceeds  in  stock  at  $104  per  share. 
How  many  shares  did  he  purchase  ?  Ans.  lb  sliares. 

2.  Three  daughters,  Mary,  Jane,  and  Ellen,  are  to  share  an 
estate  of  180000,  in  the  proportion  of  J,  \,  and  J-,  respectively ; 
but  Ellen  dies,  and  the  whole  amount  is  to  be  divided  in  a 
proper  proportion  between  the  other  two.  What  share  does 
each  receive  ?  Ans.  Mary,  $48,000 ;  Jane,  $32,000. 

3.  What  must  be  the  dimensions  of  a  rectangular  bin  that 
will  hold  350  bushels  of  grain,  if  its  length  is  twice  its  width, 
and  its  width  twice  its  depth  ? 

Ans.  Length,  15.1G+fL;  width,  7.58+ ft.;  depth,  3.71)  +  ft. 

4.  A  Chicago  merchant  bought  800  barrels  of  flour  at  $7 
per  barrel,  and  sent  it  to  New  York,  paying  9^  of  the  cost  for 
freight  and  other  charges;  his  agent  sold  it  at  an  advance  of 
25;^  on  the  original  cost  and  charged  d%  commission.  What 
was  the  net  gain  ?  Ans.  $686. 

5.  What  sum  invested  in  railroad  stock  paying  7^  annually 
will  yield  a  quarterly  dividend  of  $325.50  ?      Ans.  $18,600. 

6.  A,  B,  and  C  together  c^  dig  a  ditch  in  4  days.  A  can 
dig  it  alone  in  10  days ;  B  can  dig  it  alone  in  12  days.  How 
long  will  it  take  C  to  do  the  work  alone  ?        Ans.  15  days. 

7.  A  person  owning  7i  acres  in  the  form  of  a  rectangb 
3  times  as  long  as  it  is  wide,  wishes  to  tether  his  horse  to  a 
stake  by  the  shortest  rope  that  will  allow  him  to  graze  upon 
any  part  of  the  field.    What  is  the  length  of  rope  required  ? 

Ans.  31.62+  rd. 

8.  A  cubical  block  contains  64  cubic  feet ;  what  is  the  dis- 
tance from  one  corner  to  the  opposite  diagonal  corner  ? 

Ans.  6.92+  feefc. 

9.  A  farmer  bought  a  horse,  wagon,  and  plow  for  $134; 
the  horse  cost  ^  as  much  as  the  wagon,  and  the  plow  \  as 
much  as  the  horse.    What  was  the  cost  of  each  ? 

Ans,  Horse,  $70;  wagon,  $50;  plow,  $14. 
(356) 


REVIEW    EXAMPLES.  009 

10.  A  grocer  mixed  15  pounds  of  Hyson  tea  with  9  pounds  of 
Gunpowder  tea,  and  sold  it  at  $.95  per  pound,  thus  gaining 
2b%  on  the  original  cost.  If  a  pound  of  the  Gunpowder  cost 
16  cents  more  than  a  pound  of  the  Hyson,  what  was  the  cost 
of  each  per  pound?      A7is,  Gunpowder,  $.86  ;  Hyson,  8.70. 

11.  A  farmer  has  a  cornfield  whose  width  is  to  its  length  as 
3  to  4,  and  contains  4f  acres.  The  hills  of  corn,  supposing 
them  to  occupy  only  a  mathematical  point,  are  3  feet  apart, 
and  no  hill  is  nearer  the  fence  than  3  feet.  What  must  he  pay 
a  man  to  hoe  his  corn,  at  the  rate  of  $.50  per  day,  if  he  hoes 
750  hills  in  a  day?  Ans.  $34.23-rV 

12.  The  duty  at  20^  ad  valorem  on  a  quantity  of  tea  in 
chests,  each  weighing  75  pounds  gross,  and  invoiced  at  $.70 
per  pound,  was  $6,552,  tare  being  4^.  How  many  chests  were 
imported?  Ans.  650. 

13.  A  room  is  22  feet  long,  18  feet  wide,  and  14  feet  high. 
What  is  the  distance  from  one  of  the  lower  corners  to  tho 
opposite  upper  corner  ?  Ans,  31.68 -f  ft. 

14.  A  farmer  sold  85  sheep  at  $2,  $2.20  and  $2.80  per  head, 
and  thus  realized  an  average  price  of  $2.40  per  head.  What 
number  of  each  did  the  lot  contain  ? 

Ans.  17  at  $2;  34  at  $2.20;  34  at  $2.80. 

15.  If  the  ratio  of  increase  of  a  certain  crop  is  3,  and  a  man 
begins  by  planting  5  bushels,  using  all  the  crop  for  seed  the 
next  year,  and  so  on ;  what  will  be  his  crop  the  seventh  year  ? 

Ans,  10,935  bushels. 

16.  A  can  do  a  piece  of  work  in  4J  days  that  requires  B 
6  days  and  C  9  days  to  do  the  same  amount  of  work.  In  how 
many  days  can  they  do  it  working  together  ?     Ans.  "i  days. 

17.  A  father  divided  his  property  among  his  wife  and  four 
sons,  directing  that  his  wife  should  have  $8  as  often  as  the 
oldest  son  $6,  the  second  son  $3  as  often  as  the  wife  $5,  the 
youngest  son  $12  as  often  as  the  third  $14,  the  third  son  $5 
as  often  as  the  oldest  $7.  The  youngest  son  received  $4,500 ; 
what  was  the  value  of  the  father's  property?  Ans,  $32,780. 

(357) 


230  REVIEW    AND     TEST    EXAMPLES. 

18.  If  72  men  dig  a  trench  20  yd.  long,  1  ft.  G  in.  hroad, 
and  4  ft.  deep  in  3  days  of  10  hours  each,  how  many  men 
would  be  required  to  dig  a  trench  30  yd.  long,  2  ft.  3  in.  broad, 
and  5  ft.  deep  in  15  days  of  9  hours  each  ?       Ans.  45  men. 

19.  Samuel  Wells  paid  3 J  times  as  much  for  a  house  as  for 
a  barn  ;  had  the  barn  cost  him  Q%  more,  and  the  house  S% 
more,  the  whole  cost  would  have  been  $7260.  What  was  the 
actual  cost  ?  Ans.  $6,750. 

20.  Cliange   ^  of —  to  a  simple  fraction,  and  re- 

1  J i_ 

'^3  +  i 
duce  to  lowest  terms.  Ans,  ^. 

21.  A  person  sells  out  $4,500  of  4:%  stock  at  95,  and  invests 
the  proceeds  in  bank  stock  at  80,  which  pays  an  annual  divi- 
dend of  2^%.     How  much  is  the  gain  or  loss  per  annum  ? 

Ans.  $37.50  loss. 

22.  James  Griswold  bought  f  of  a  ship ;  but  the  property 
having  fallen  in  value  8%,  he  sells  14;^  gf  his  share  for  $2700. 
What  was  the  value  of  the  ship  at  first  ?  Ans.  $25,000. 

23.  A  gentleman  willed  to  the  youngest  of  his  five  sons 
$2000,  to  the  next  a  sum  greater  by  one  half,  and  so  on,  the 
oldest  receiving  $10,125,  thus  disposing  of  his  entire  estate. 
What  was  the  gentleman  worth  ?  Ans.  $26,375. 

24.  I  sent  $7847  to  my  agent  in  New  Orleans,  who  pur- 
chased sugar  at  an  average  price  of  $16  per  barrel ;  he  charged 
3i%  commission.    How  many  barrels  did  he  buy?    Ans.  475. 

25.  Bought  3,000  bushels  of  wheat  at  $1.50  per  bushel. 
What  must  I  ask  per  bushel  that  I  may  fall  20%  on  the  asking 
price  and  still  make  16%,  allowing  10%  of  the  sales  for  bad 
debts?  Ans.  $2.41|. 

26.  Henry  Swift  has  $6,000  worth  of  5%  stock;  but  not 

being  satisfied  with  his  income,  he  sells  at  96  and  invests  in 

stock  paying  4^%,  which  gives  him  an  income  greater  by 

$45.60.     At  what  price  did  he  purchase  the  latter  stock  ? 

Ans.  At  75, 
(358) 


REVIEW    AND     TEST    EXAMPLES.  331 

27.  A  drover  bought  a  number  of  horses,  cows,  and  sheep 
for  $3,900.  For  every  horse  he  paid  $75,  for  each  cow  he  paid 
I  as  much  as  for  a  horse,  and  for  each  sheep  \  as  much  as  for 
a  cow.  He  bought  3  times  as  many  sheep  as  cows,  and  twice 
as  many  cows  as  horses  ;  how  many  did  he  buy  of  each  ? 

Ans.  20  horses;  40  cows;  120  sheep. 

28.  I  shipped  to  my  agent  in  Buffalo  a  quantity  of  flour, 
which  he  immediately  sold  at  $7.50  per  barrel.  I  then  in- 
structed him  to  purchase  goods  for  me  at  a  commission  of 
3 J^ ;  he  charged  me  4^  commission  for  selling,  and  received 
as  his  whole  commission  $800.  How  many  barrels  of  flour  did 
I  send  him  ?  Ans,  1,472. 

29.  Adam  Gesner  gave  his  note  for  11,250,  and  at  the  end 
of  3  years  4  months,  and  21  days,  paid  off  the  note,  which  then 
amounted  to  81504.375 ;  reckoning  only  simple  interest,  what 
was  the  rate  ^?  A71S.  6^. 

30.  A  hound  in  pursuit  of  a  fox  runs  5  rods  while  the  fox 
runs  3  rods,  but  the  fox  had  60  rods  the  start.  How  far  must 
the  hound  run  before  he  overtakes  the  fox?    Ans.  150  rods. 

31.  A  man  divided  his  property,  amounting  to  $15,000, 
among  his  three  sons,  in  such  a  manner  that  their  shares  put 
at  Q%  simple  interest  should  all  amount  to  the  same  sum  when 
they  were  21  years  old ;  the  ages  of  the  children  were  respec- 
tively 6  yr.,  9  yr.,  and  13  yr.     What  was  the  share  of  each  ? 

Ans.  Oldest,  $5683.082+  ;  second,  $4890.094+; 
youngest,  $4426.822  +  . 

32.  A  certain  garden  is  12f  rods  long,  and  9 J  rods  wide. 
At  2|-  cents  per  cubic  foot,  what  will  it  cost  to  dig  a  ditch 
around  outside  it  that  shall  be  3J  feet  wide  and  4  feet  deep  ? 

Ans.  $258.03|. 

33.  A  farmer  sells  a  merchant  40  bushels  of  oats  at  $.60  per 
bushel  and  makes  20^  ;  the  merchant  sells  the  farmer  4  yards 
of  broadcloth  at  $3.75  per  yard,  15  yards  of  calico  at  8  cents 
per  yard,  and  40  yards  of  cotton  cloth  at  12  cents  per  yard, 

(359) 


23S  REVIEW    AND     TEST    EXAMPLES, 

and  makes  a  profit  of  25%,    Which  gains  the  more  by  the 
trade  and  how  much  ?  Ans.  Merchant  gains  $.20. 

34.  A  triangular  cornfield  consisting  of  146  rows,  has  437 
hills  in  the  longest  row,  and  2  in  the  shortest;  how  many 
corn  hills  in  the  field  ?  Ans,  32047  hills. 


35.  What  is  the  value  of 


r44f  of  .056  —  3.04  of  ^~| 
|_     (8_2.4)+f  of3|     J 


(Si^of^l-l 


:76  +  ^* 


^n-. 

+  2 


285, 
^551^ 


A71S.  a, 

36.  Bought  60  barrels  of  flour  at  18.50  per  barrel,  but  on 
account  of  its  having  been  damaged,  one-half  of  it  was  sold  at 
a  loss  of  10^,  and  the  remainder  at  19  per  barrel.  What  %  was 
lost  by  the  operation  ?  Ans.  2^%, 

37.  A  room  22  feet  long,  16  feet  wide,  and  9  feet  high,  con- 
tains 4  windows,  each  of  which  is  5 J-  feet  high  and  3  feet  wide; 
also  two  doors,  7  feet  in  height  and  3|-  feet  in  width.  The 
base-boards  are  |  of  a  foot  wide.  What  will  it  cost  to  plaster 
and  paper  the  room,  if  the  plastering  cost  16  cents  per  square 
yard  and  the  papering  10  cents  ?  A7ts.  $21.20jij. 

38.  Two  persons,  A  and  B,  each  receive  the  same  salary. 
A  spends  76^^  of  his  money,  and  B  spends  as  much  as  would 
equal  45J^  of  what  both  received.  At  the  end  of  the  year 
they  both  together  have  left  $276.25 ;  what  part  of  it  belongs 
to  A,  and  what  to  B?  Ans.  A,  $199.75;  B,  $76.50. 

39.  Two  persons  280  miles  apart  travel  tow^ard  each  other 
until  they  meet,  one  at  the  rate  of  6  miles  per  hour,  the  other 
at  the  rate  of  8  miles  per  hour.   How  far  does  each  travel  ? 

Ans.  First,  120  miles;  second,  160  miles. 

40.  James  Welch  has  a  debt  in  Chicago  amounting  to 
$4489.32.  For  what  sum  must  a  note  be  drawn  at  90  days,  that 
when  discounted  at  6%  at  a  Chicago  bank,  will  just  pay  the 
debt  ?  Ans,  $4560. 


REVIEW    AND     TEST    EXABTPLES.  233 

41.  I  went  to  the  store  to  buy  carpeting,  and  found  that 
any  one  of  three  pieces,  width  respectively  1},  1|,  and  2 J 
yards,  would  exactly  fit  my  room  without  cutting  anything 
from  the  width  of  the  carpet.    What  is  the  width  of  my  room? 

Ans.  22|-  feet. 

42.  If  10  horses  in  25  days  consume  3^  tons  of  hay,  how 
long  will  ^  tons  last  6  horses,  12  cows,  and  8  sheep,  if  each 
cow  consumes  |  as  much  as  a  horse,  and  each  sheep  f  as  much 
as  a  cow?  Ans.  25  days. 

43.  The  distance  between  the  opposite  corners  of  a  square 
field  is  60  rods  ;  how  many  acres  in  the  field  ? 

Ans.  11  A.  40  sq.  rd. 

44.  At  $225  per  ton,  what  is  the  cost  of  17  cwt.  2  qr.  21  lb. 
of  sugar?  Ans.  $199.237J. 

45.  A  drover  bought  12  sheep  at  $6  per  head ;  how  many 
must  he  buy  at  $9  and  $15  per  head,  that  he  may  sell  them 
all  at  $12  per  head  and  lose  nothing  ? 

Ans.  1  at  $9,  25  at  $15. 
4G.  Three  men  bought  afield  of  grain  in  circular  form  con- 
taining 9  A.,  for  which  they  paid  $192,  of  which  the  first  man 
paid  $48,  the  second  $64,  the  third  $80.  They  agreed  to  take 
their  shares  in  the  form  of  rings ;  the  first  man  mowing  around 
the  field  until  he  got  his  share,  then  the  second,  and  so  on. 
What  depth  of  ring  must  each  man  mow  to  get  his  share  of 
the  grain  ?  Ans,  1st  man,  2.86  +  rd. ; 

2d  man,  4.73+  rd.; 

3d  man,  13.81+  rd. 

47.  A  young  man  inherited  an  estate  and  spent  15^  of  it 
during  the  first  year,  and  30^  of  the  remainder  during  the 
second  year,  when  he  had  only  $9401  left.  How  much  money 
did  he  inherit  ?  Ans.  $15800. 

48.  Mr.  Webster  bought  a  house  for  $6750,  on  a  credit  of 
10  months ;  after  keeping  it  for  4  months,  he  sold  it  for  $7000 
on  a  credit  of  8  months.  Money  being  worth  Q%,  what  was 
his  net  cash  gain  at  the  time  of  the  sale  ?    Ans,  $177.37  +  . 

(361) 


234  BE  VIEW    AND     TEST    EXAMPLES, 

49.  A  and  B  can  do  a  piece  of  work  in  18  days;  A  can  do 
f  as  much  as  B.   In  how  many  days  can  each  do  it  alone  ? 

Ans.  A,  40^  days  ;  B,  32|  days. 

60.  If  -I  of  a  farm  is  worth  $7524  at  $45  per  acre,  how  many 
acres  in  the  whole  farm  ?  Ans.  195^  A. 

51.  A  person  paid  $1450  for  two  building  lots,  the  price  of 
one  being  45^  that  of  the  other ;  he  sold  the  cheaper  lot  at  a 
gain  of  60^,  and  the  dearer  one  at  a  loss  of  2b%,  What  %  did 
he  gain  or  lose  on  the  whole  transaction  ?    Ans.  1-^%  gain. 

52.  A  certain  sum  of  money,  at  %%  compound  interest  for 
10  years,  amounted  to  $2072.568.  What  was  the  amount  at 
interest.?  Ans.  $960. 

53.  There  are  two  church  towers,  one  120  feet  high,  and 
the  other  150  feet.  A  certain  object  upon  the  ground  between 
them  is  125  feet  from  the  top  of  the  first  and  160  feet  from 
the  top  of  the  second ;  how  far  apart  are  their  tops  ? 

Ans.  95.50-f  feet. 

54.  A  farmer  sold  to  a  merchant  80  bushels  of  wheat  at 
$1.90  per  bushel,  70  bushels  of  barley  at  $1.10,  and  175  bushels 
of  oats  at  $.75.  He  took  in  payment  a  note  for  5  months,  and 
immediately  got  it  discounted  at  bank  at  6^;  how  much 
money  did  he  receive  ?  Ans.  $351,064-. 

55.  There  is  a  pile  of  100  railroad  ties,  which  a  man  is 
required  to  carry,  one  by  one,  and  place  in  their  proper  places, 
3  feet  apart ;  supposing  the  first  to  be  laid  3  feet  from  the 
pile,  how  far  will  the  man  travel  in  placing  them  all  ? 

A71S.  30300  feet. 

56.  Sound  travels  at  the  rate  of  1142  feet  a  second.  If  a 
gun  be  discharged  at  a  distance  of  4J  miles,  how  much  time 
will  elapse,  after  seeing  the  flash,  before  the  report  is  heard  ? 

Ans.  20|4J  sec. 

57.  If  a  company  of  480  men  have  provisions  for  8  months, 
how  many  men  must  be  sent  away  at  the  end  of  6  months, 
that  the  remaining  provisions  may  last  6  months  longer  ? 

Ans.  320  men. 
(362) 


REVIEW    AND     TEST    EXAMPLES.  235 

58.  The  first  year  a  man  was  in  business  he  cleared  $300, 
and  each  year  his  profit  increased  by  a  common  difference ; 
the  fourteenth  year  he  made  $950.  How  much  did  he  make 
the  third  year  ?  Ans,  $400. 

59.  What  number  is  that,  which  being  increased  by  i,  \, 
and  f  of  itself,  and  diminished  by  25,  equals  291  ? 

Ans.  180. 

60.  At  what  time  between  5  and  6  will  the  hour  and 
minute  hands  of  a  clock  be  together  ? 

Ans.  27^  min.  past  five. 
CI.  A  field  whose  length  is  to  its  width  as  4  to  3  contains 
2  A.  2  R.  32  rd. ;  what  are  its  dimensions  ? 

Ans.  Length,  24  rd. ;  width,  18  rd. 

62.  Three  persons  formed  a  partnership  with  a  capital  of 
$4600.  The  first  man's  stock  was  in  trade  8  months  and 
gained  $752 ;  the  second  man's  stock  was  in  trade  12  months, 
and  gained  $600 ;  and  the  third  man  had  his  stock  in  16 
months,  and  gained  $640.    What  was  each  man's  stock  ? 

A71S.  First,  $2350;  second,  $1250;  third,  $1000. 

63.  How  many  thousand  shingles,  18  inches  long  and  4  in. 
wide,  lying  ^  to  the  weather,  are  required  to  shingle  the  roof 
of  a  building  54  feet  long,  with  rafters  22  feet  long,  the  first 
row  of  shingles  being  double  ?  Ans.  14|f . 

64.  Employed  an  agent  who  charges  4:%  commission  to  col- 
lect a  bill  of  $550.  He  succeeded  in  obtaining  only  85^;  how 
much  did  I  receive  ?  Ans.  $44S.S0. 

65.  A  and  B  entered  into  partnership  and  gained  $4450.50. 
A  put  in  enough  capital  to  make  his  gain  15^  more  than  B's ; 
what  was  each  man's  share  of  the  gain  ? 

A71S.  A,  $2380.50  ;  B,  $2070. 

66.  A  building  is  75  feet  long  and  44  feet  wide,  and  the 
elevation  of  the  roof  is  14  feet.  How  many  feet  of  boards  will 
be  required  to  cover  the  roof,  if  the  rafters  extend  2  feet 
beyond  the  plates,  and  the  boarding  projects  IJ  feet  at  each 
end,  and  ^  allowed  for  waste  ?  Ans,  5474.97  +  feet, 

(363) 


236  REVIEW    AND     TEST    EXAMPLES. 

67.  A  circular  court  is  laid  with  19  rows  of  flat  stones,  each 
row  forming  a  complete  circle ;  the  outside  row  is  39  inches 
wide,  and  the  width  of  each  row  diminishes  2  inches  as  it 
nears  the  centre.    What  is  the  width  of  the  innermost  row  ? 

Ans,  3  inches. 

■I  of  4  of  4A 

68.  Reduce   ?— ^J— ^-^  to  a  simple  fraction,  and  take  the 

^  of  I  of  ff  ^ 

result  from  the  sum  of  lOf,  3^^,  and  7||.  Ans.  3f  J. 

69.  Bought  75  yards  of  cloth  at  10^  less  than  the  first  cost, 
and  sold  it  at  10^  more  than  the  first  cost  and  gained  $25. 
What  was  the  first  cost  per  yard?  Ans.  81.0Gf, 

70.  A  grain  merchant  bought  7500  bushels  of  corn  at  81.35 
per  bushel,  5450  bushels  of  oats  at  $.80,  3250  bushels  of  barley 
at  $.95,  paid  8225  for  freight  and  $170  for  storage;  he  imme- 
diately sold  it  an  advance  of  20^  on  the  entire  cost,  on  a  credit 
of  6  months.  What  %  did  he  gain  at  the  time  of  the  sale, 
money  being  Avorth  8^  ?  Ans.  15 -\-%. 

71.  A  farmer  employs  a  number  of  men  and  8  boys ;  he  pays 
the  boys  $.65  and  the  men  $1.10  per  day.  The  amount  that  he 
paid  to  all  was  as  much  as  if  each  had  received  $.92  per  day; 
how  many  men  were  employed?  Ans.  12  men. 

72.  S.  Howard  can  mow  6  acres  in  4  days,  and  his  son  can 
mow  7  acres  in  5  days.  How  long  will  it  take  them  both  to 
mow  494  acres  ?  Ans.  17^  days. 

73.  I  lent  a  friend  $875,  which  he  kept  1  year  and  4  months. 
Some  time  afterward  I  borrowed  of  him  $350;  how  long 
must  I  keep  it  to  balance  the  favor  ?  Ans.  3  yr.  4  mo. 

74.  Find  the  difierence  between  the  surface  of  a  floor  80  ft. 
9  in.  long  and  65  ft.  6  in.  broad,  and  the  sum  of  the  surfaces 
of  three  others,  the  dimensions  of  each  of  which  are  exactly 
one-third  of  those  of  the  other. 

Ans.  391  sq.  yd.  7  sq.  ft.  12  sq.  in. 

75.  A  tree  broken  off  24  feet  from  the  ground  rests  on  the 
stump,  the  top  touching  the  ground  30  feet  from  the  foot  of 
the  tree.    What  was  the  height  of  the  tree  ?    Ans.  62.41  -j-  ft 

(364) 


BHVIEW    AND     TEST    EXAMPLES.  237 

76.  Two  persons  entered  into  partnership  for  trading.  A 
put  in  '^^245  for  375  days  and  received  -|  of  the  gain  ;  the  num- 
ber of  dollars  that  B  put  m  was  equal  to  the  number  of  days 
it  was  employed  in  trade;     Wlmt  was  B's  capital  ? 

Ans.  $350. 

77.  How  many  square  feet  of  boards  1^  inches  thick  will  be 
required  to  make  a  box,  open  at  the  top,  whose  inner  dimen- 
sions are  6  feet  long,  4  feet  wide,  and  3  feet  deep  ? 

Ans.  88-^  sq.  ft. 

78.  A  farmer  having  80  acres  of  land,  worth  $55  an  acre, 
wishes  to  buy  enough  more  at  ^50  and  165,  respectively,  so 
that  the  value  of  his  land  shall  average  $60  an  acre.  How 
much  of  each  must  he  buy  ?    Ans.  1  A.  at  $50  ;  82  A.  at  $65. 

79.  What  is  the  amount  of  an  annuity  of  $700  for  8  years, 
at  Q'/o  compound  interest  ?  Ans.  $6928.22 1. 

80.  What  must  be  the  price  of  stock  yielding  ^^%,  that  will 
yield  the  same  profit  as  4^^  stock  at  90  ?  Ans.  112. 

81.  A  person  after  spending  \  and  J  of  his  money  and  $20, 
had  $80  left.     What  had  he  at  first  ?  Ans.  $240. 

82.  James  Harper  has  a  large  jewelry  store,  which  with  its 
contents  he  insures  in  the  Continc^ntal  Insurance  Company 
for  f  of  its  estimated  value,  at  3J;^.  This  Company  imme- 
diately insures  J  of  its  risk  in  the  Astor  Company,  at  2^%, 
After  two  years  and  a  half,  the  store  and  its  contents  were 
destroyed  by  fire,  when  it  was  found  that  the  Astor  Company 
lost  $2925  more  than  the  Continental  Company.  Reckoning 
(j%  simple  interest  on  the  premiums  that  the  owner  paid,  what 
would  be  his  entire  loss  ?  A71S.  $78815.75. 

83.  A  drover  sold  42  cows  and  34  oxen  for  $3374,  receiving 
$21  per  head  more  for  the  oxen  than  for  the  cows.  What  did 
he  receive  for  each  per  head  ?  Ans.  $35  for  cows ; 

$56  for  oxen. 

84.  A  certain  room  is  27  ft.  5  in.  long,  14  ft.  7  in.  wide, 
and  12  ft.  10  in.  high.  How  much  paper  |  of  a  yard  wide  will 
be  required  to  cover  the  walls  ?  Ans,  136  yd.  ^  ft,  8  in. 

(365) 


2^  REVIEW    AND    TEST    EXAMPLES. 

85.  The  area  of  a  triangular  field  is  6  A.  36  rd. ;  the  base 
is  64  rods.  What  is  the  perpendicular  distance  from  the  base 
to  the  angle  opposite?  Ans.  31^  rods. 

86.  If  the  width  of  a  building  is  50  feet,  and  the  length  of 
the  rafters  30  feet,  what  will  it  cost  to  board  the  gable  ends, 
at  $.18  per  square  yard  ?  Ans.  $16.58  +  . 

87.  What  is  the  solidity  of  the  largest  ball  that  can  be  cut 
out  of  a  cubical  block  whose  sides  are  6  inches  square  ? 

Ans.  113.0976  cu.  in. 

88.  A  privateer  took  a  prize  worth  £348  15s.,  which  was  to 
be  divided  among  1  captain,  3  mates,  and  27  privates,  so  that 
a  private  should  have  one  share,  a  mate  twice  as  much  as  a 
private,  and  the  captain  6  times  as  much  as  a  mate.  What 
was  the  share  of  each  ? 

Ans,  Private,  £7  15s. ;  mate,  £15  10s.;  captain,  £93. 

89.  The  width  of  a  certain  building  is  38  feet,  and  the  eleva- 
tion of  the  roof  is  16  feet ;  how  many  square  feet  of  boards 
will  be  required  to  cover  the  gable  ends  ?       Ans.  608  sq.  ft. 

90.  The  length  of  one  side  of  a  field  in  the  form  of  an  equi- 
latei-al  triangle  is  40  rods.  How  many  acres  does  the  field 
contain,  and  what  would  it  cost  to  fence  it,  at  $.  65  per  rod  ? 

Ans,  4  A.  52.8+  sq.  rd.;.  178. 

1  +  1  +  1 

2  3       4 

91.  Change  ^ ^ z-  to  a  simple  fraction,  and  reduce 

to  lowest  terms.  Ans.  1|-J. 

92.  Two  merchants,  Sanford  and  Otis,  invested  equal  sums 
in  trade.  Sanford  gained  a  sum  equal  to  J  of  his  stock  and 
$24  more,  and  Otis  lost  $444 ;  then  Otis  had  just  J  as  much 
money  as  Sanford.     What  did  each  invest?  A^is.  $675. 

93.  Henry  Norton  sold  his  farm  for  $13270,  $5000  of  which 
was  to  be  paid  in  6  months,  $4000  in  one  year  and  6  months, 
and  the  rest  in  2  years.  What  was  the  net  cash  value  of  his 
fann,  money  being  worth  Q%  ?  Ans.  $12336.59+. 

(366) 


REVIEW    AM)     TEST    EXAMPLES,  239 

94.  At  what  time  between  10  and  11  o'clock  will  the  hands 
be  directly  opposite  ?  Ans.  21^  min.  past  ten. 

95.  How  much  better  is  it  to  invest  $15000  in  6%  stock,  at 
a  discount  of  25^,  than  to  let  the  same  sum  at  7^  simple 
mterest?  Ans.  $150. 

96.  What  is  the  present  worth  of  an  annuity  of  $550  for 
6  years,  at  8%  simple  interest?  Ans.  $2675.075 -f- . 

97.  What  is  the  value  of  (t^i±^^)  ^  1^  ? 

\         5  —  4f         /       .005 

Ans.  i^, 

98.  If  36  men  working  8  hours  a  day  for  16  days  can  dig 
a  trench  72  yd.  long,  18  ft.  wide,  and  12  ft  deep,  in  how  many 
days  will  32  men,  working  12  hours  a  day,  dig  a  trench  64  yd. 
long,  27  ft.  wide,  and  18  ft.  deep  ?  Ans.  24  days. 

99.  Bought  a  piece  of  broadcloth  at  $2.75  per  yard.  At 
what  price  shall  it  be  marked  that  I  may  sell  it  at  h%  less 
than  the  marked  price  and  still  make  20^  profit  ? 

Ans.  $3.47yV 

100.  A  man  hired  a  mechanic  for  35  days,  on  condition  that 
for  every  day  he  worked  he  should  receive  $1.75,  and  for  every 
day  he  was  absent  he  should  forfeit  $2.50.  At  the  end  of  the 
time  he  received  $40 ;  how  many  days  did  he  work  ? 

Ans,  30  days. 

101.  If  stock  bought  at  25^  premium  pay  1\%  on  the  invest- 
ment, what  %  will  it  pay  if  bought  at  4^  discount  ? 

Ans.  ^%. 

102.  The  interest  on  a  note  for  2  yr.  3  mo.  18  da.,  at  8%,  was 
$155.02  ;  what  was  the  face  of  the  note  ?         Ans.  $842.50. 

103.  The  distance  on  the  road  around  a  certain  park  is  17 
miles.  If  three  persons  start  from  the  same  point  on  the  road 
at  the  same  time  and  travel  in  the  same  direction  around  the 
park,  how  far  will  each  have  to  travel  before  they  all  come 
together,  if  the  first  travels  5  miles  an  hour,  the  second  6, 
and  the  third  7  miles  an  hour? 

Ans.  First,  85  mi. ;  second,  102  mi. ;  third,  119  mi. 
(367) 


240  REVIEW    AND     TEST    EXAMPLES. 

104.  Two  persons  commence  trade  with  the  same  amount  of 
money ;  the  first  man  spends  4:8%  of  his  yearly,  and  the  second 
spends  a  sum  equal  to  25%  of  what  both  had  at  first ;  at  the 
end  of  the  year  they  both  together  had  $3468.  How  much 
had  each  at  the  end  of  the  year  ? 

Ans.  $1768,  first;  $1700,  second. 

105.  A  roller  used  for  leveling  a  lawn  being  6  ft.  6  in.  in 
circumference  by  2  ft.  3  in.  in  width,  is  observed  to  make  12 
revolutions  as  it  rolls  from  one  extremity  of  the  lawn  to  the 
other.  Find  the  area  rolled  when  the  roller  has  passed  ten 
times  the  whole  length  of  it.  A?is.  195  sq.  yd. 

106.  A  and  B  form  a  copartnership :  A's  stock  is  to  B's  as  5 
to  7.  At  the  end  of  4  months  A  withdraws  f  of  his  stock,  and 
B  J  of  his ;  their  year's  gain  is  $5650.  How  much  does  each 
receive?  Ans.  A,  $2500  ;  B,  $3150. 

107.  A  drover  bought  a  number  of  sheep,  oxen,  and  cows. 
He  paid  half  as  much  more  for  oxen  as  for  sheep,  and  half  as 
much  more  for  cows  as  for  oxen  ;  he  sold  the  sheep  at  a 
profit  of  10^,  the  oxen  at  a  profit  of  S%,  and  the  cows  at  a 
loss  of  4:% ;  he  received  for  the  whole  $3416.  What  did  he 
pay  for  each  lot  ? 

A71S.  $700,  sheep ;  $1050,  oxen ;  $1575,  cows. 

108.  A  commission  merchant,  who  charges  If^,  purchases 
for  me  145  barrels  of  sugar,  pays  for  freight  $12.50,  making 
the  whole  bill  $2255.07.  If  there  were  190  pounds  of  sugar 
in  each  barrel,  what  was  the  price  of  the  sugar  per  pound,  and 
what  was  the  amount  of  commission  ? 

Ans.  $.08  per  pound;  $38.57,  com. 

109.  A  merchant  in  New  York  imported  from  England  a 
quantity  of  goods,  for  which  he  had  to  pay  a  duty  of  12^.  On 
account  of  the  depression  in  trade,  he  is  obliged  to  sell  at  a 
loss  of  7f  ^ ;  had  he  sold  them  two  months  sooner,  he  would 
have  received  $896  more  than  he  did,  and  then  would  have 
cleared  di%  on  the  transaction.  What  price  did  he  pay  for 
the  goods?  Ans.  $7500. 

(368) 


REVIEW    AND     TEST    EXAMPLES.  241 

110.  How  many  bricks  8  inches  long,  4  inches  wide,  and 
2  inches  thick  will  be  required  to  build  a  cubical  cistern,  open 
at  the  top,  that  shall  contain  2000  gallons,  if  the  wall  is  made 
a  foot  thick  and  \  of  the  entire  wall  is  mortar  ? 

Ans,  5918  bricks- 
Ill.  What  is  the  value  of  ^  x  — ^.-  ?         Ans.  |||. 

34  +  rj 

112.  If  18  men,  working  10  hours  per  day,  can  dig  a  ditch 
in  20  days,  how  long  will  it  take  3  men  and  40  boys,  working 
8  hours  per  day,  to  dig  a  ditch  twice  as  long,  6  men  being 
equal  to  10  boys  ?  Ans,  33-J-  days. 

113.  John  Turner  owes  $350  due  in  7  months,  $500  in  3 
months,  and  $650  due  in  5  months,  and  pays  -f  of  the  whole 
in  G  months;  when  ought  the  remainder  to  be  paid? 

Ans,  3  months. 

114.  A  and  B  can  do  a  piece  of  work  in  4f  days;  B  and  0 
in  5^  days ;  and  A  and  C  in  4f  days.  In  what  time  can  each 
do  the  work  alone?   Ans,  A,  8  days;  B,  10  days;  C,  12  da. 

115.  A  and  B  aJone  can  do  a  piece  of  work  in  15  and  18 
days  respectively.  They  work  together  on  it  for  3  days,  when 
B  leaves ;  but  A  continues,  and  after  3  days  is  joined  by  C ; 
together  they  finish  it  in  4  days.  In  what  time  could  C  do 
the  piece  of  work  by  himself  ?  Ans,  24  days. 

116.  Mr.  Smith  paid  3 J  times  as  much  for  a  horse  as  for  a 
harness.  If  he  had  paid  10^  less  for  the  harness  and  1l\% 
more  for  the  horse,  they  would  together  have  cost  $245.40. 
How  much  did  he  give  for  each  ? 

Ans.  Horse,  $182 ;  harness,  $56. 

117.  James  and  Herbert  are  running  around  a  block  25  rods 
square ;  James  runs  around  it  every  7^  minutes,  and  Herbert 
every  S-J  minutes.  If  they  started  together  from  the  same 
point,  how  many  times  must  each  run  around  the  block 
before  they  will  be  together  ? 

Ans,  James,  10  times ;  Herbert,  9  times. 
{369) 


t4&  REVIEW    AND    TEST    EXAMPLES. 

118.  James  Walker  contracted  to  build  a  stone  wall  180  rd. 
long  in  21  days.  He  employed  45  men  12  days,  who  built 
41 2 J  yards.  How  many  more  men  must  be  employed  to  finish 
the  work  in  the  required  time  ?  Ans.  39. 

119.  I  invested  $6345  in  Government  bonds  at  104J,  bro- 
kerage 1^%.  How  much  would  I  gain  by  selling  the  same  at 
1134,  brokerage  1^%?  Ans.  ^375. 

120.  A  grain  merchant  bought  3250  bushels  of  wheat,  at 
$1.25  per  bushel,  and  sold  it  immediately  at  $1.45  per  bushel, 
receiving  in  payment  a  note  due  4  months  hence,  which  he 
had  discounted"  at  bank  at  6%.    What  was  his  gain  ? 

Ans.  $553.39  +  . 

121.  Two  men  form  a  partnership  for  trading ;  A's  capital 
is  13500,  B's  14800.  At  the  end  of  7  months,  how  much  must 
A  put  in  that  he  may  receive  J  of  the  yeai**s  gain  ? 

A71S.  $3120. 

122.  A  man  having  lost  25^  of  his  capital,  is  worth  exactly 
as  much  as  another  who  has  just  gained  15;^  on  his  capital ; 
the  second  man's  capital  was  originally  19000.  What  was  the 
first  man's  capital  ?  Ans.  $13800. 

123.  A  merchant  imported  18  barrels  of  syrup,  each  con- 
taining 42  gallons,  invoiced  at  $.95  per  gallon ;  paid  $85  for 
freight  and  a  duty  of  30^.  What  %  will  he  gain  by  seUing 
the  whole  for  $1171.459  ?  Ans.  \b%. 


Art.  307. 

O      12.10.18.'^.     9 

•^'   t8  »    tF>  ^8  >    28 '  ^^' 

/r      13  .    30.   A.  A.    _Z. 

7.  3if. 

^-  FT- 

9.  $5272  jV 

12.  f. 
i.?.  49. 
i4-  $1614|. 
15.  40|  tons, 
ie.  121 1  yards. 
11.  $63 1. 
i^.  2H. 

19.  %im\. 

20.  47i-;irai.;  38^|  mi- 

21.  $8113  rV 

22.  556,%. 

23.  $10500. 
2J^.  504. 
;?5.  $1800. 

26.  123 1  cords. 
;^7.  4i|f  pounds ; 

Sj^'V  pounds ; 

f  I  pounds. 
j?5.  Increased  by  4i. 

29.  %imh 

30.  2}4  days. 

ol-  S65HfM. 
^^.  15  feet. 

33.  1196|fayds.;  %%i-^. 

34.  $289||. 

35.  13.V.I  acres. 
^<?.  $5006^. 


Art.  451. 

1.  38.1024  Kg. 

2.  33.5627  Ton. 
^.  33.8304  HI. 

4.  2.2185  L. 

5.  68.0194  St. 
(?.  13274.16  A. 

7.  250  cu.  in. 

8.  38  sq.  yd. 

5.  7.7353+  bu. 

10.  552960  gr. 

11.  36.8965+  cd. 

12.  1176.15+  cu.  ft. 

13.  $.024624. 

14.  $1.41975. 

15.  $14.5675  +  . 

16.  $1,357  +  . 

17.  8340  dg.  ;  83.4  Dg. 

18.  8400  L.  ;  840000  cl, 

19.  790.75+  mi. 

20.  I960. 

21.  $74.00048  +  . 

Art.  467. 

1.  8400. 

2.  76000. 

3.  573. 

4.  38097. 

5.  897.52. 

6.  30084. 

7.  3426000. 

8.  720000000. 

9.  463000000. 

Art.  468. 

1.  300800. 

2.  18018. 

3.  350000. 

4.  1673800. 

5.  24473.6. 


^.  11500000. 
7.  27360000. 
5.  414000. 
9.   3141. 

Alt.  469. 

i.  6628122. 

2.  385605. 

5.  87546366. 

4.  53111520. 

5.  839516040. 

6.  2585632. 

7.  7349272. 

8.  62767170. 

9.  4388206. 

Alt.  470, 

1.  87.36. 

2.  43.72. 

3.  7.903. 

4.  .02397. 

5.  .05236. 

6.  .0006934 

7.  .0054. 
5.  .00007. 
9.   .0072. 

Art.  471. 

7.  183.8. 

2.  6.54625. 

5.  .1098^ 

4.  .013. 

5.  .008. 

6.  .0004. 

7.  .674f. 
.?.  2.7041?. 


Art.  472, 


244 


ANSWERS. 


W 


3. 

4- 
5. 
6.  A. 

JO.  tItf. 

n.  i. 
IB.  ^Uj, 

Art.  473. 


'*•   TTJ- 

3.  If. 

4.  30. 

5.  J. 

I'  ¥^' 

8.  7. 

9.n. 

10.  30. 
Ji.  130. 
12.  2331. 

Art.  474. 

1.  28. 
;^.  30. 
^.  103. 
J^.  25. 

^:  ffe. 

7.  158f. 
^.  80. 
S.  50. 

Art.  479. 

1.  15300. 

2.  48700. 
;?.  596|. 
//.  111875. 
5.  78000. 
C.  9655|. 
7.  8000. 
^.  14.6. 

S.  1090. 
to.  1472& 


i5.  $1112; 
$1251 ; 
$1042.50; 
$973. 

17.  $425;  $550; 
$612.50  ; 
$533i. 

18.  $600;  $315; 
$910  ;  $700. 

Id.   $630;  $675; 
$650. 

20.  $980;  $1470;! 
$735. 

21.  $112  ;  $128. 
$149,331. 

22.  $1110. 

23.  $371.25; 
$309.371 ; 
$433. 12|. 

2lt.   $255  ;  $510 ; 
$408. 

25.  $548  ; 
$657.60  , 
$1315.20 ; 
$1534.40. 

26.  $522; 
$543.75 ; 
$696; 
$761.25. 

Art.  480. 

1.  217.74. 

2.  2619. 

3.  1952.8. 

4.  5025. 
6.   11.388, 

6.  1.9708. 

7.  3.1584. 

8.  1.8434. 

9.  3.62552. 

10.  219.288. 
25.  358.16  A. ; 

179.08  A. ; 
268.62  A. ; 
8.954  A. ; 
71  632  A.; 
537.24  A. ; 
35.816  A. 
16,  5083.2  bu.  ; 
2541.6  bu. 

11.  7926  yd. ; 


18494  yd. ; 
10568  yd.  ; 
792.6  yd. ; 
1056.8  yd. ; 
2377.8  yd. ; 
1849.4  vd. 

18.  $1725:' 
$197l|  ; 
$2300 ; 
$3450 ; 
$1863 ; 
$2116 ; 
$2380.50. 

Art.  494. 

10.  $56. 

11.  $87. 

12.  $144,441. 

13.  $35.57^. 
U.  $143. 46f. 

15.  $17.79f. 

16.  $4656.88|. 
n.  $5.68ii. 
IS.  $52.66f. 

19.  $19.11yV 

20.  $940.95. 

Art.  495. 

6.  $.38f|. 

7.  $1.4825. 

8.  $78.66  +  . 

9.  $4,375. 

10.  $2.75. 

11.  $5758  +  . 

12.  $15.50. 

Art.  496. 

1,  $4.05. 

2,  $7812; 
$12361 H. 

3,  $668.25  ; 
$767.88 ; 
$816.48. 

J^.  $1163.35. 
5.  $465.75  ; 
$465.75. 
e.  $3.4545. 

7.  $1846.2295 

8.  $273.12525. 

9.  $379.64024. 


Art.  497. 

5.  274|  yd. 

6.  525  bu. ; 
h\^%  bu.  ; 
630  bu. ; 
534/v  t>u. ; 
458j\  ^u. 

7.  51  rd. 

8.  217  A. 

9.  285  lb. 

10.  395  yd. 

11,  476  bu. 


Art.  498. 

{.  30  yd. 

?.  37  T.  ; 


3. 


24T8rV  T. 
38|  bu. ; 
40  bu. 


Jt.  96  ft. 

5.  483  yd. 

6.  89. 

7.  53  cd. 
5.  9  ft. 

Art.  499. 


7.  $585.50. 

8,  $120.25. 
5.  $5075. 

10.  $847.09|. 

11.  $40455. 

Art.  500. 

^.  f ;  f;  t;i. 

7       71 

8.  i8f. 

9.  4t^V. 

2.^.  II,  or  .68. 
13.  i,"||. 

^5.  H;*. 


ANSWERS. 

24 

Art.  601. 

22.  550. 

U.  $4000. 

15.  $384,048+. 

8.  42 ;  84. 

^e?.  10  yd. 

16.  $596. 

329;  905|. 

2I^.  8. 

Art.  515. 

17.  $24.60. 

S.  20.4  ;  77.76  ; 

^5.  $35500. 

1.  $3.04. 

i<?.  $375,435. 

504. 

^6'.  $1.50. 

^.  $81.60. 

i5.  112  bbl. 

4.  $60; 

27.  3100  lb. 

J.  $74.48. 

20.  $13.82. 

S641.35I  ; 

^<9.  $75. 

4.  |l58.2312i. 

^2.  $129,375. 

$1.35A. 

29.  1. 

5.  $116.57^.  " 

22.  $6761.882+, 

5.  22.90.  " 

5^.  300  A. 

6.  $2278.96.1. 

€.  $58.88; 

31.  $75. 

7.  15%. 

Art.  527. 

15.80  lb. 

5^.  $1350. 

8.  2dj\%. 

1.  $105. 

7.  78^  A.; 

S3.  $83331. 

9.  20%. 

2.  $199.37i. 

32W. 

.?4.  $412500. 

10.  $6355.362. 

3.  $148.58|. 

8.  1.96  yd. 

^5.  $8000. 

11.  $.78. 

4.  2|-%. 

9.  15  men. 

12.  $77.40  gain; 

5.  $227111-, 

10.  25.02  lb. 

Art.  505. 

$6.30  selling 

6.  $6000. 

li.  13.25i  mi. 

7.  25%. 

price. 

7.  1%. 

12.  91.50. 

8.  16|%. 

9.  20^. 

13.  $8.25. 

5.  $250.66f. 

i.?.  $6.45. 

U  $7.14. 

9.  $1948.80. 

U.  $837.2. 

i^.  n\%. 

15.  $5,121. 

10.  $148.75. 

15.  S347.75. 

11.  291%. 

i5.  $5860. 

J6\  24i  lb. 

12.  16\fo. 

17.  $9.50  buying 

Art.  540. 

17.  120.7  yd. 

13.  441%. 

price  ; 

2.  $8965.50. 

IS.  259.86  A. 

U'  0||%. 

$7.12i  selling 

3.  $10105.621-. 

19.  $1173 ; 

i5.  35f%; 

price. 

4.  76  shares. 

$4.25  per  yd. 

5?%; 

18.  $9.75. 

5.  96  shares. 

SO.  84  ct. 

10f%. 

19.  $1825. 

6.  144  shares. 

16.  42H%. 

f^.  $570.24. 

7.  153  shares. 

Art.  503. 

17,  n-u%. 

18.  26|%. 

^i.  20%. 

8.  $15680. 

1.  108. 

22.  $3475. 

5.  $40090. 

S.  $45 ;  $03. 

i9.  76^\%. 

^5.  $153. 

10.  $16462.50. 

S.  324  vd. 

20.  39H%. 

24.  $3.20  per  yd. 

11.  $59800. 

4.  $1323. 

;?/.  8%. 

^5.  $4,221  per  cd. 

12.  $70. 

/J.  $23580. 

22.  UiUfo. 

26.  20%." 

7,?.  $62.50. 

23.  Q\\\%. 

;^7.  $.76^. 

74.  $58,331-. 
15.  $128.5t|. 

Art.  503. 

2J^.  6im%. 

25.  33i%. 

7.  200. 

Art.  522. 

i6".  $392. 

5.  400. 

^6.  12i%. 

1.  $14,131-. 

17.  $850.20. 

9.  1200. 

27.  10%. 

^.  $18.2105. 

i<§.  $448.47. 

i^.  800. 

28.  15|%. 

3.  $57.75. 

i.9.  $583.05. 

i7.  2400  vd. 

^^-  nm\%' 

4.  $96.95952. 

20.  $451  in. 

12.  700  bu. 

5.  21%. 

21.  $964. 

n.  20. 

Art.  507. 

e.  41%. 

22.  $284,982+. 

i4.  $9.H. 

e.  216. 

7.  2f  %. 

23.  $368. 

i5.  m  pk. 

7.  184. 

8.  $14400. 

^4.  $1160. 

iC.  14  ft. 

5.  560  mi. 

5.  $47178.12i. 

25.  281  shares. 

17.  21. 

9.  872. 

10.  $1652.92/;. 

^^.  $1029. 

!<?.  4. 

7^.  800  men. 

11.  $8663.21-,V:;. 

27.  6%. 

i£>.  42f. 

11.  $1000. 

12.  $3653.70  +  . 

^.9.  $191.80. 

€0.  $36. 

i^.  $6.40. 

13.  $1863.97+. 

29.  $3720. 

^1.  87^. 

i5.  $2000. 

U-  $3286. 

J(?.  Uif  %. 

24G 


ANSWERS. 


51.  175. 

82.  |62i. 

53.  $53." 

54.  $5950. 

56.  $680. 

^6.  $800,821+  ; 
$78.^  82+  ; 
$124,931  +  . 

57.  $8. 

Art.  554. 

1.  $268  80. 
J?.  $255,192. 

5.  $622. 68f. 

4.  $73.60. 

6.  $66.6792. 

Art.  566. 

13.  $51.5256. 

U-  $282.33. 

15.  $8.3695. 

16.  $41.78265. 

17.  $86.3208  +  . 
IS.  $1.6559  +  . 
19.  $13.25248. 
SO.  $107.1144. 
81.  S85.115. 

52.  $827.08. 

83.  $462,616. 
^4'  $42362. 
85.  $5,736. 
S6.  $97.1694. 

Art.  570. 

1,  $63,048; 
$89,318. 

5.  $113.1074; 
$145.4238. 

5.  $118.3442; 
$73,965  +  . 

/f.  $64.1775; 
$106.9625. 

6.  $815,976+; 
$1078.254  +  , 

6.  $292.3719. 

7.  $49,529  +  . 

8.  $1094.096. 

9.  $410,475  +  . 
10.  |1699.b0  +  . 


Art.  573. 

1.  $35.84. 

2.  $48,675. 

3.  S43.812. 
J,.  $12,754 
5.  $28.1885. 
G.  $35.82. 

7.  $120. 

8.  $46.50. 

9.  $40.20. 
10.  $100,395. 


Art.  576. 

1.  $1.58. 

2.  $1,536. 

3.  $2,125. 

4.  $4.2075. 

$1.54f. 

$1.849i 

$2.67f. 

$3.7754. 

$1,122. 


6. 
6. 
7. 
8 
9. 
10.  $1.96. 

Art.  578. 

1.  $62.36352; 
$93.54528 ; 
$83.15136. 

2.  $303.9513; 
$434.216 ; 
$173.6875. 

S.  $22.2609  ; 
$11.1304 ; 
$19.4783. 

4.  $113.40; 
.'^151.20; 
$56.70. 

5.  $45.4765; 
$57.7202; 
$66.4656. 

Art.  582. 

1.  $24.65; 
$39.44; 
$34.51. 

2.  $62.6533 
$93.9800; 
$41.7689. 

3.  $291,695; 
$458,378; 
$120.Om. 


4. 


$103.44004 

$131.2892 ; 

$57.6877. 
■>.  $65.9458; 

$50.3270 ; 

$19.0895. 
>.  $1.85; 

$2.6037 ; 

$l.r3074. 
y.  $376.6183 ; 

$502.1577; 

$313.8486. 

Art.  585. 

t.  $11.5436. 
$3.1342. 
$3.3082. 
$3.1574. 
$3.3945. 
$3.6073. 
$6.54407. 
$1.4576. 
$16.2754. 
$.8939. 
$461,193. 


12.  $25.2125. 
Art.  588. 


2.  $364.9937  +  . 

3.  $283,992  +  . 

4.  $462,019. 

5.  $562,984. 
G.  $296. 

7.  $434,994. 

Art.  591. 

2.  $264,998+. 

3.  $49,652  +  . 

4.  $295,996  +  . 

5.  $572,996  +  . 

Art.  594. 

2.  Qfc. 

3.  7%. 

4.  Sfo. 
5.5%. 

6.  8^%  nearly. 

7.  12%. 

5.  ^%  better  2d. 
0.  14|%, 


10.  25%  I  12^%. 
]nfo  ;  S^%  ; 
4%. 

11.  33i%;  50%; 
22|%;33^%; 
14f  %;  21f  %; 
111%;  16|%. 

Art.  597. 

1.  2  yr.  3  mo. 

2.  6  yr. 

3.  3  yr.  6  mo. 

4.  4  yr.  9  mo. 

5.  2  yr.  8  mo. 

G.  5   vr.    7  mo. 
6  da. 

7.  6  yr.    4  mo. 

24  da. 

8.  7  yr.    6  mo. 

25+  da. 

9.  14f  yr. 

10.  20  yr. ; 
12iyr.; 
15^yr.; 
ll|^yr.;40yr.; 
25yr.  ; 

'SOU  yr.  ; 
22|  yr. 

11.  71|yr. 

Art.  601. 

1.  $146.27795. 

2.  $977,532. 

3.  $31390106+. 

4.  $1854.576. 

5.  !5^6o4.5102  +  . 

6.  $20.0034. 

7.  $205,516  +  . 

8.  $108,595  +  . 

9.  $44.0824 +gr. 
at  comp.  int. 

Art.  604. 

1.  1219.558. 

2.  $183.0183. 

3.  $133.55053. 

4.  $11  4.8721673^ 

5.  $55  4364. 

C.  $99.2659725, 


AJVSWURS. 


247 


Art.  606. 

1.   $469.53704. 
^.  $1230.2528. 

3.  $781.52013. 

4.  $2755.3006. 

5.  $506.8663. 

6.  $348.1372193. 

7.  $357.399443. 

8.  $328.3345. 

9.  $145.7:>8068. 
20.  $556.75033. 
11.   $753.052567. 
li\  83439.63075. 

13.  $1097.5152. 

14.  $854.943736. 

Art.  609. 

1.  $146,004. 
^.  $1071.41i. 
3.  $1138.075. 
4-  $33.1858. 

5.  $363,308. 

6.  $50.56  dif.  between 

Sim.  and  Annual 
Int.  ; 

$4,309  dif.  between 
An.  and  Com.  Int ; 

$51,769+  dif.  be- 
tween Sim.  and 
Com.  Int. 

Art.  616. 

1.  $283.46}. 
£.  $11,254. 
3.  $302,793. 
4-  $117,942. 

Art.  619. 

e,  $1491.49  +  . 

3.  $2891.5^7. 

4.  $420,293. 
6.  $5434.651  +  . 

Art.  624. 

i.  $776,699+; 

$754,717  +  . 

e.  $315,789+  ; 

$348,387  +  . 

5.  $485,468+  ; 
$478.10+. 

4  $9  975  +  . 


$148,456. 

5.  $5,513  ; 
$40.83  +  . 

6.  $15,275; 
$230.57R. 

7.  $530,367. 

8.  $9171.90  +  . 

9.  $.957. 

10.  $435. 

11.  $.103  more   profit- 

able at  $4.66. 
1^.  3d  $33  865  better. 
13.  $1.47  +  . 

Art.  632. 

1.  $5.88  Bk.  Dis. ; 
$374.13  Proceeds ; 
$11,941  Bk.  Dis.  ; 
$368.05i  Proceeds. 

2.  $30.6711  Bk.  Dis. ; 
$769.3381,  Proc'ds; 
$11.530|Bk.  Dis.  ; 
$778.479.V  Proc'ds. 

3.  $35.83f  Bk.  Dis.  ; 
$1574. 17f  Proc'ds 
$54.033f  Bk.  Dis. 
$1545.9771  Proc'ds 

4.  $.884. 

5.  $30.13U. 

6.  $37,194. 

7.  Due  May  37; 
59  da.  Time  ; 
$475,383+  Proc'ds 

8.  Due  Aug.  16  ; 
75  da.  Time ; 
$581.3951  Proc'ds. 

9.  Due  Aug.  22  ; 
91  da.  Time  ; 
$1571.681  Proc'ds. 

Art.  635. 

1.  $876,061  +  ; 
$295,415  +  ; 
$540713. 

^.   $458,387  ; 


4.  $480,616. 

5.  $354,453. 

6.  $398,899  +  . 

7.  $961,781,  1st; 
$967,495,  3d; 
$979,914,  3d. 

8.  $495,363. 

9.  $1517.440. 

Art.  643. 

1.   $3416. 


$99,113. 
$238.63  ; 
$1830.923; 
$515,648. 


3.  $850.68. 

4.  $6331.50. 

5.  $133,796. 

6.  $1491. 

7.  $382.3038*. 

8.  $391,141. 

9.  $486. 

10.  $560. 

11.  $730. 

12.  $480. 

13.  $824 
U.   $375.50. 

15.  $321. 

16.  $403. 

17.  $698.25. 

18.  $1615.11*-. 

19.  $2415.925. 
^0.  $1779. 

21.  $496. 

22.  97TV'%,or3U%dia 

23.  $3536.38^. 

24.  $8013.43  +  . 

Art.  G48. 

2.  $2134.99065  +  . 

3.  $81)3.22?. 

4.  $1(>42.41. 
6.  $763. 

6.  $3469. 33k 

7.  $609,375. 

8.  $11456.8125. 

Art.  650. 

1.  $12615.96. 

2.  $301  gain  by  Ind. 

3.  $124852+  less  by 
Ind. 

4.  3011.58+  marks. 


248 


ANSWERS. 


Art.  661. 

1.  July  1, 1876. 
^.  Dec.  27,  1876. 

3.  April  30,  1876. 

4.  Oct.  It),  1876. 
^.  Sept.  3,  1876. 
C.   July  21,  1877. 
7.  Oct.  19, 1877. 

5.  60  da. 
i?.  Feb.  5. 

Art.  664. 

1.  $210,  Face  of  note ; 

Due  Dec.  12,  1876. 
^.  $100  due; 

Dec.  7, 1876,  equat- 
ed time. 
S,  Apr.  6,  1877. 
4.  March  1, 1876. 


Art.  681. 


i.  f. 

0     14» 
"•    3TTK- 


4.  //s- 

5.  «gW. 


6.  Y- 


Art.  683. 


5. 

C.    y. 

7.  V- 

s.h 

9.  jV 

Art.  685. 

/.  150  da. 
S.  45  A. 
;.'.  $597. 

4.  $2100. 

5.  $2386.40. 

Art.  687. 

1.  104  A. 

e.  $50. 

^.  130  da. 


4.  $fi78|f. 

5.  415  lines. 

6.  177  cd.  3  cd.  ft. 

7.  $2000. 

<9.  17  gal.  3  qt.  1  pt. 
9.  $13.50. 

Art.  690» 

2.  $72. 
5.  $468. 
^.  $27. 

5.  $100. 

6.  $204. 

7.  403|. 

5.  228f  yd. 

Art.  700. 

1.  35. 
^.  6. 
S.  272. 

4.  29^  bu. 

5.  10  yd. 

C.  £168  15s.  6f€L 

1.  56. 

2.  21. 
^.  15. 

5.  213^. 
<?.  8  cwt. 

Art.  702. 

4.  195  bu. 

5.  $9. 

6.  70  bu. 

7.  803  ft. 
<?.  $18. 

9.  $83,655. 

10.  $3000. 

11.  $7. 
li?.  $2.40. 

75.  16  mi.  109  rd. 

14.  160  gal.  3  qt.  1  pt. 

15.  $.98x=V. 

16.  2  yr.  10  mo. 

17.  ^5^375. 

Art.  70a 


i.  48. 

2.  23|. 

Art.  706. 

i.  264. 

2.  120  cd. 

5.  180  bu. 

4.  $18. 

5.  $116760. 

6.  12  lb. 

7.  1728  ft. 

8.  23040  yd. 

Art.  711. 


2.  $405,125. 
S.  $3222.261. 

4.  $300  for  4  mo. 

5.  $2400  for  4  mo. 

Art.  713. 

1.  $1661.538, 

A's  sliare ; 
$1107.692. 

B's  share; 
$1550.769, 

C's  share. 

2.  12382.545, 

A's  share; 
$1737.272, 

B's  share; 
$1340.181, 

C's  share.  - 

3.  $2814.128, 

A's  share ; 
$1644.112, 

B's  share ; 
$3431.758, 

C's  share  ; 

4.  $1178.947,  ^tna ; 
$1547.368,  Home; 
$2394.736,  MutuaL 

5.  $900, 1st  district ; 
$700,  2d  district ; 
$600,  3d  district ; 
$400,  4th  district. 

Art.  715. 

1.  $210,  A's  share; 
$144,  B's  share, 

2,  $70,451,  A's  share; 


AJVS  WEB  Si 


249 


$39,629,  B's  share  ; 

$26,419,  C's  share. 

5,  $4940,  A's  stock  ; 
$P>510,  B's  stock  ; 
$7150,  C's  stock. 

4.  P35.10  +  , 

A's  profit ; 
^288.56  +  , 

B's  profit ; 
$251.32  +  , 

C's  profit. 

6.  $3666.06  +  , 

A's  share ; 

$5238.93  +  , 

B's  share. 

Art.  717. 

5.  $1.00. 

5,  $.13^V. 

4.  18  J  carats. 

6.  $.311. 

Art.  722. 

1.  1,  3,  2,  1  lb. 

5.  3  gal.  of  mo.  to  3 

gal.  of  water. 

5.  1  part  of  each. 
4.  1, 2.  and  6  bbl. 

6.  2,   2,  604,  and  240 

bbl. 

7.  2,  1,  and  109  gal. 

8.  50,   50,    5,    and    1 


10.  18, 27, 63,  and  27  lb. 

11.  '30,  30,  and  180  oz. 

12.  44|,89t,and89|lb. 

Art.  728. 

8.  2304:  4225,186624. 

5.  86436;  148996; 
247009 ;  64009. 

4.  250047 ;  15625  ; 
438976;  60236288. 

.512. 

6.  56169. 

7.  4100625. 

O        2197 


Q         289 

10.  .00390625. 

11.  .039304 


13.   .00028561. 
U.   .000000166375. 
15.   .00091204. 

Art.  729. 

2.  7. 

3.  7. 

4.  9. 

5.  15. 

6.  12. 

7.  13. 
5.  12. 

iO.  30. 
11.  24. 
ii?.  12. 

Art.  745. 

1.  64.    7.  If. 

2.  59.     5.  If. 

3.  53.     5.  If. 

4.  87.    10.   .15. 

5.  93.    11.   .48. 
G.  96.    i;^.  .76. 

13.   371. 
i4.  64.5. 
15.  876.6. 
i6.  167.4. 
i7.  7.56. 
IS.   21.79. 
i9  5.656+. 

20.  7.681 +. 

21.  2.645+. 
j^^.  .964  +  . 
23.   .894+. 
^4.  .612  +  . 
25.   3.834+. 
;?e.  9.284+. 
^7.  2.443+. 
28.   .881+. 
;?«?.  .346  +  . 
30.   .404  +  . 
,?i.  51. 

^^.  Q\ 
S3.  33587. 
J4.  6. 
35.  2656. 
56'.  354.906 +  . 
57.  95  ft. 

38.  487  ft. 

39.  69.57+  yd. 
<^.  96  trees. 
41.  44.342+  rd. 


4^.  1.4142;  2.2360; 

3.3166. 
43.  .654;. 852;  .735. 
Art.  753. 

1.  6.  9.  M. 

^.  9.        m  j.f. 

5.  11.  i7.  63. 

4.  13.  i^.  76. 

5.  16.  13.  289. 

6.  22.  14.  361. 

7.  19.  i5.  .y^. 
<?.  W.  i6.  4.84. 

17.  2.22+;  3.30 +  ; 

4.37 +;  6.17  +  ;. 56+ ; 

.75:4.22  +  ;  1.88 +. 
IS.  1.442+ ;  1.912+ ; 

.793  +  ;  .341  +  ; 

.208  +  ;  1.273 +. 

19.  4. 

20.  37. 
^/.  9. 

'.  76.36  +  . 
■.  8. 

24.  439  ft. 

25.  78.3+  inches, 
^(7.  2730f  sq.  ft. 

27.  32  feet. 

28.  196.9+  inches. 
i'».  436  feet. 

30.  26.9+  feet. 
Art.  765. 
1.  26. 
^.  13  in. 

3.  10. 

4.  70;  433. 

5.  $18. 
e.  12. 

7.  500500. 
5.  $97.60. 

9.  12  days ;  336  mi. 
10.  144 

Art.  709. 
1.  49152. 
;^.  2048. 
5.  7^ 

4.  2m. 

5.  393213. 

6.  315  mi. 

7.  189  mi. ;  & 
^.  3;  4378. 


tm^ 


ANSWERS. 


Art.  775. 

1.  $1410. 

2.  $1905. 

3.  $2485.714  +  . 

4.  $1775.9772. 

6.  $491.73  +  . 
€.  $5288.88+. 

7.  $8944.226. 

8.  $2380.59+. 

9.  6%. 

10.  7%. 

11.  $500. 

12.  $2500. 

13.  3  years. 

Art.  799, 

1.  53}  sq.  ft. 

2.  3/r  sq.  rd. 

3.  108  sq.  cli. 

4.  112.292  sq.  ch. 
^.  510  sq.  ft. 

G.  21  GO  stones ; 
$307.50. 

Art.  800, 

1.  10  ft. 

2.  22  rd. 

3.  16  ft.  8  in. 
Jf.  42  rd. 

5.  6  yd. 

C.  4  rd.  4  yd.  1  ft.  9  in. 

7.  20  ft. 

Art.  801. 
1.   150  sq.  ft. 

8.  612.37  sq.  in. 
8.   .924+  A. 

4.  692.82+  sq.  ft. 

5.  $292.68  +  . 

Art.  802. 

1.  43.08+  ft. 

2.  39  ft. 

S.   128.80+  ft. 

4.  187.45+  rd. 

5.  11.14+  rd. 

Art.  803. 

1.  37.08+  ft. 

2.  40.31+  ft. 
5.  2.10+  ft. 

4.  42  ft. 

5,  18  ft. 


Art.  804. 

2.  28  sq.  ft.  108  sq.  in.; 
6  sq.  yd.  6  sq.  ft. 

138  sq.  in. 
^.  7  A.  3  sq.  ch.  15  P. 
125  sq.  1. 

3.  2805  sq.  ft. 

4.  29  sq.  ft. 

Alt.  805. 

1.  220  sq.  ft. 

2.  57  sq.  in. 

5.  44  sq.  ft. 
4.   $161.25. 

Art.  806. 

1.  450  sq.  in. 

2.  104  sq.  ft. 

3.  300  sq.  ft. 

4.  11  A. 

Art.  807. 

1.  30  in.  ;  25  ft. 

2.  43.9824  in. ; 
56.5488  in. 

3.  $45,289  +  . 

4.  596904000  mi. 

Art.  808. 

1.  314.16  sq.  ft.  ; 
1385.4456  sq.  in. ; 
19()3.50  sq.  ft. 

2.  14313.915  sq.ft.; 
3911.085  sq.  ft.; 
$17.55  +  . 

3.  79.5727+  A. 

Art.  809. 

1.  8  ft. 

2.  43.9324  ft. 

3.  5  rd. 

4.  37.6902  in.  . 

6.  62.832  rd. 
6.   16  in. 

Art.  8*0. 

1.  84  sq.  ft. 

2.  36  ,^  gq.  ft. 

3.  135  sq.  in. 

4.  180  sq.  ft. 

6.  826.7204  sq.  ft. 

Art.  820. 

1.  103.8  cu.  ft. 

2,  411.5132+  cu.  ft. 


3.  1800  cu.  ft. 

4.  238.7610  cu.  ft 

5.  $31.25. 

6.  41.888  cu.  ft. 

Art.  821. 

1.  114  sq.  in. 

2.  235.62  sq.  ft. 

3.  35  sq.  ft. 

4.  323  sq.  ft. 

5.  731  sq.  ft. 

6.  519.58+  sq.  in. 

7.  738.276  sq.  ft. 
S.   608.685  sq.  in. 

Art.  822. 

1.  25.1328  cu.  ft. 

2.  50.2C56  cu.  ft. 

3.  80  cu.  ft. 

4.  166.27+  cu.  in. 

5.  3600  cu.  ft. 

Art.  823. 

1.  45  sq.  ft. 

2.  135  sq.  in. 

3.  337.29+  sq.  ft. 

4.  1124.6928  sq.  ft. 

Art.  824. 

1.  1184  cu.  ft. 

2.  348.7176  cu.  ft. 

3.  24.871  cu.  ft. 

4.  48631.968  cu.  ft. 

5.  9445.448+  cu.  ft 

Art.  825. 

1.  201.0624  sq.  ft. 

2.  254.4696  sq.  in. 

3.  6361.74  sq.ft.. 

4.  4071.5136  sq.  in* 

I  5.   78.54  sq.  in. 

Art.  820. 

1.  4188.8  cu.  in. 

2.  14.1372  cu.  yd. 

3.  14137.2  cu.  in. 

4.  .22089  cu  ft. 

5.  268.0832  cu.  in. 

Art.  827. 

1.  58.752  gal. 

2.  104.1012  gal. 

3.  52.6592  gaL 

4.  85  gal. 


^B  35828 


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